HA150 Hana
January 11, 2017 | Author: Sumeet Patil | Category: N/A
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HA150 SQL Basics for HANA
Material Number: 50119271
SQL Basics For HANA – Agenda Motivation And Basic Concepts Reading Data From A Table Or View
Exercises 1 - 19
Aggregating Data
Exercises 20 - 29
Reading Data From Multiple Tables Part I
Exercises 30 - 40
Reading Data From Multiple Tables Part II
Exercises 41 - 45
Understanding NULL Values Changing Data Stored in Tables
Exercises 46 - 53
Defining How Data Is Stored
Exercises 54 - 58
Using Views For Data Access
Exercises 59 - 60
Defining Data Access Database Transactions ©
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SQL Basics For HANA Course
The SQL Basics course is designed to ...
Explain basic concepts in the database world and the Relational Database Model
Refresh and deepen SQL knowledge, especially for developing on HANA database systems
The SQL Basics course is not designed to ...
Explain implementation details of SAP HANA
Supersede official SAP HANA documentation
Solve specific development problems
Serve as a basis for decision-making for development projects
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Unit 1 Motivation & Basic Concepts
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Unit 1: Motivation And Basic Concepts Learning Objectives After completing this unit, you will be able to: • List some database models • Understand the motivation for and foundation of the relational model • Understand why learning SQL is important when dealing with HANA • Understand how the relation model and SQL are related • List the three sub-languages of SQL • Understand the sample database used throughout the course
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5
Basic Concepts
What are the terms •
Database
•
Database System
•
Database Management System ?
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DBS = DB + DBMS
The Database Management System (DBMS) manages the database
Every access to the database (create, read, insert, update, delete) goes exclusively through the DBMS The DBMS exercises complete control over the database
Database (DB) = structured collection of "records"
Database System (DBS) = specific database + DBMS
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DBS = DB + DBMS
User / Application
DBMS DBS
DB ©
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DBS = DB + DBMS
„What Database do you use?“
„Which Database Management System do you use?“
„We are using HANA as Database.“
„We are using SAP HANA as Database Management System.“
In everyday life, these terms are usually used incorrectly …
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Basics Concepts
3 Level Schema Architecture
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3 Level Schema Architecture
Objective: Changes at a lower level should not affect a higher level (if possible)
Application/User
External Schema 1
Application/User …
External Schema n
Conceptual Schema
Internal Schema
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External Level
Conceptual Level
Internal Level
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3 Level Schema Architecture
The address book should not display salary data External Level (VIEW) An Employee has a D-Number, name, and salary and is assigned to a department Conceptual Level (TABLE) The board mostly accesses employee data according to descending order of salary (which has to be very fast) Internal Level (INDEX)
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Basic Concepts
Which Database Models are available?
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Database Models
Hierarchical Database Model
Network Database Model
Relational Database Model
Object-relational Database Model
Object-oriented Database Model
XML-based Database Model
…
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Basic Concepts
Relational Database Model
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Relational Database Model
Invented in the late 1960s by Edgar F. Codd (1923-2003) IBM Almaden Research Lab in San Jose (California)
First prototype in the mid-1970s • "System R", IBM San Jose
Relational DBMS products (selection) • • • •
©
Oracle (since 1979) IBM DB2 (since 1983) SQL Server (since 1989) SAP HANA (since 2011)
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Relational Database Model
Initial Development Goals (1960s): Simple but mathematically profound database model – Easy to understand but still mathematically precise
Simple but mathematically-based database language – Easy to learn, but semantics still described with mathematical precision – Provable equivalence of two queries
Integrity monitoring largely by the DBMS Storage & retrieval of data is the responsibility of the DBMS (and not of the user) – Descriptive language rather than navigation (optimizer selects optimal execution strategy)
Clean separation between conceptual and internal schema
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Basic Concepts
Why is the Relational Database Model actually called Relational Database Model?
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Relations
A relation (in the sense of mathematics) is the subset of the Cartesian product of sets
R⊆AxBxC
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Comparative Relations (=,≠,,≥)
N = {1, 2, 3, 4, 5} N × N = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5)}
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Comparative Relations (=,≠,,≥) ≤ ⊆ N × N ≤ ⊆ {(1,1), (2,1), (3,1), (4,1), (5,1),
(1,2), (2,2), (3,2), (4,2), (5,2),
(1,3), (2,3), (3,3), (4,3), (5,3),
(1,4), (2,4), (3,4), (4,4), (5,4),
(1,5), (2,5), (3,5), (4,5), (5,5)}
≤ = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,2), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5), (4,4), (4,5), (5,5)} (2,3) ϵ ≤ 2 ≤ 3
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Relations PersNumber = {P036407, P040824, P052867, …} Name = {Ben, Paul, Raja, …} HiringYear = {2001, 2008, 2010, …} PersNumber × Name × HiringYear= {(P036407,Ben,2001), (P036407,Ben,2008), (P036407,Paul,2001), (P036407,Paul,2008), (P035607,Raja,2001), (P036407,Raja,2008), (P040824,Ben,2001), (P040824,Ben,2008), (P040824,Paul,2001), (P040824,Paul,2008), (P040824,Raja,2001), (P040824,Raja,2008), (P052867,Ben,2001), (P052867,Jim,2008), (P052867,Paul,2001), (P052867,Vishal,2008), (P052867,Raja,2001), (P052867,Werner,2008),
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(P036407,Ben,2010), (P036407,Paul,2010), (P036407,Raja,2010), (P040824,Ben,2010), (P040824,Paul,2010), (P040824,Raja,2010), (P052867,Ben,2010), (P052867,Paul,2010), (P052867,Raja,2010), …}
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Relations Employee ⊆ PersNumber × Name × HiringYear Employee ⊆ {(P036407,Ben,2001), (P036407,Paul,2001), (P035607,Raja,2001), (P040824,Ben,2001), (P040824,Paul,2001), (P040824,Raja,2001), (P052867,Ben,2001), (P052867,Paul,2001), (P052867,Raja,2001),
(P036407,Ben,2008), (P036407,Paul,2008), (P036407,Raja,2008), (P040824,Ben,2008), (P040824,Paul,2008), (P040824,Raja,2008), (P052867,Jim,2008), (P052867,Vishal,2008), (P052867,Werner,2008),
(P036407,Ben,2010), (P036407,Paul,2010), (P036407,Raja,2010), (P040824,Ben,2010), (P040824,Paul,2010), (P040824,Raja,2010), (P052867,Ben,2010), (P052867,Paul,2010), (P052867,Raja,2010), …}
Employee= {(P036407,Paul,2001), (P040824,Ben,2008), (P052867,Raja,2010)}
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Relations & Tables
A relation can be represented as a table Employee= {(P036407,Paul,2001), (P040824,Ben,2008), (P052867,Raja,2010)}
Employee
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PersNumber
Name
HiringYear
P036407 P040824 P052867
Paul Ben Raja
2001 2008 2010
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Basic Concepts
Which languages are available for the Relational Database Model?
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Relational Languages
Relational Algebra
Formal basis for DBMS internal query optimization 6 basic operations (selection, projection, cross join, union, difference, rename)
Relational Calculus
Tuple variables and quantifiers
SQL
©
Standardized and used in practice
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Why SQL?
Why SQL is important?
Why it is worth working with SQL
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Why SQL? → Reasons #1 and #2 In almost every business application scenario, the data is managed using database systems. The most significant are database systems based on the relational data model and using SQL (Structured Query Language) as a database language.
SQL is a widely-established, powerful, standardized database language many application programmers have experience in. There is (so far) no other database language that has all the advantages mentioned.
SAP HANA is a relational data base management system and SAP HANA supports SQL ©
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Why SQL?
Different application architectures and development models are possible with SAP HANA Data Marts with SAP HANA
Standalone HANA Apps
BI Tools, MS Excel, Web Browser …
Standalone Application SAP HANA clients
Traditional DBMS
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HANA Content / Database Tables
HANA Content / Database Tables
SAP HANA
Schema replication
SAP HANA
BI 4.0 (optional) Semantic Layer
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Why SQL?
Different application architectures and development models are possible with SAP HANA HANA as primary Database HANA as Accelerator (secondary DB) (for AS ABAP) SAP GUI, Browser-based GUI
SAP GUI, Browser-based GUI
SAP Application Server
SAP Application Server
replication
Data
SAP HANA
ABAP Schema
ABAP Schema / Database Tables
SAP HANA
E.g. CRM on HANA – NW 7.40
Aggregation Levels
Traditional DBMS
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Why SQL? → Reason #3
HANA supports an extended version of SQL
No matter which HANA-based architecture or development model you work with, it is likely that using HANA SQL will be beneficial.
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SQL
SQL Features
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SQL Features
SQL is standardized – No uncontrolled growth with respect to syntax and semantics
SQL is descriptive (rather than procedural) – The "what" and not "how" is described
SQL execution is optimized – SQL statement is first parsed and optimized, and then executed – Optimizer determines optimal execution plan (at least in theory)
SQL is multi-set oriented (and not single record-based) – Using a single SQL statement multiple table rows can be read, modified or deleted in one go
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SQL Language Elements
SQL language elements can be divided into three categories:
DML = Data Manipulation Language SELECT, INSERT, UPDATE, DELETE
DDL = Data Definition Language CREATE, ALTER, DROP, RENAME
DCL = Data Control Language GRANT, REVOKE
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SQL
SQL is standardized!
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SQL Standard: History
1970s
SEQUEL (Structured English Query Language) as a language for "System R" • "System R" was the first relational prototype developed by IBM in San Jose
End of 1970
Renaming of SEQUEL into SQL (Structured Query Language)
1982
American National Standards Institute (ANSI) begins with standardization of SQL
1986
First version of the SQL standard is adopted (as ANSI) SQL-86 ("SQL0“)
1987
International Organization for Standardization (ISO) does SQL standard (ISO standard)
1989
SQL-89 ("SQL1")
1992
SQL-92 ("SQL2"), 3 attenuation levels: Entry Level, Intermediate Level, Full Level
1999
SQL:1999 ("SQL3")
2003
SQL:2003 ("SQL4"), "SQL/XML:2006" 2006 official standard part of SQL:2003
2008
SQL:2008 ("SQL5"), officially: "ISO/IEC 9075:2008" and "DIN 9075" (> 3,000 pages)
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SQL
SQL ≠ relational
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SQL
SQL is the most important database language for relational database model. Although SQL is based on the relational database model, it deviates in important points from the "purely relational" model:
No (primary) key required
NULL values are allowed
3-valued logic required ("A = A" need not be TRUE)
The language is not closed
©
Duplicates (identical rows) allowed (multi-sets instead of sets)
Order of the result rows may be relevant (ORDER BY) Anonymous result columns allowed Duplicate result column names allowed
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Database Objects
What can be found in the database?
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Database Objects
The table is the primary database object - but not the only one. Apart from tables a database usually contains:
©
Views to simplify and limit data access Indexes to speed up (certain) read accesses Constraints to ensure data consistency Stored procedures for more complex tasks Triggers to selectively respond to particular events
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Basic Concepts
What is a table?
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Tables
A (database) table represents a relation (but: multi-set of rows instead of set of tuples)
Employee PersNumber
Name
HiringYear
P036407
Paul
2001
P040824
Ben
2008
P052867
Raja
2010
Employee = {(P036407,Paul,2001), (P040824,Ben,2008), (P052867,Raja,2010)}
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Tables
A table consists of rows and columns
A table can therefore be represented two-dimensionally
A table represents a relation (with multi-sets property)
©
A table represents a (unordered) multi-set of points in n-dimensional space: Each point of the multi-set corresponds to a table row Each of the n dimensions is equivalent to a table column
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Table rows as points in space
Name
Paul HiringYear Raja
(P040824, Ben, 2008) 2010 2008
Ben
2001 PersNumber
P030407
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P040824
P052867
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Projection on the plane
Projection list
Name
SELECT PersNumber, Name FROM Employee WHERE PersNumber = 'P040824';
Paul
HiringYear Raja
(P030824, Ben, 2008) 2010 2008
Ben
(P040824, Ben)
2001 PersNumber
P036407
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P040824
P052867
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Projection on an axis
Projection list
Name
SELECT Name FROM Employee WHERE PersNumber = 'P040824';
Paul
HiringYear Raja
(P040824, Ben, 2008) 2010 2008
Ben
2001 PersNumber
P036407
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P040824
P052867
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Basic Concepts
What are the components of a (database) table?
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Tables
Table-Name
Primary Key
Column Name
Employee PersNumber Name P036407 Paul Table Row P040824 Ben P052867 Raja
HiringYear 2001 2008 2010 (Table Column) value
(Table) Column
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Basic Concepts
What is a key?
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Key
Key = is a set of columns which serves to uniquely identify any row in the table. The ability to uniquely identify rows must apply in principle (and not only for the rows existing at a certain point in time).
Employee PersNumber Name P036407 Paul P040824 Ben P052867 Raja
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HiringYear 2001 2008 2010
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Multi-column Keys
A key can consist of multiple columns
Employee Name Country City BuildingNr Block Floor Room SeatNr Mr A
DE
WDF
03
A
5
29
1
Ms B
DE
WDF
03
A
5
29
2
Ms C
DE
WDF
03
A
5
29
3
…
…
…
…
…
…
…
…
In this example the key consists of 7 columns 1 Key (not 7 keys)!
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Multiple Keys
There may be more than one key
Employee PNumber Name TaxID
Passport ID
Plate Number
ChasisNr
CarLicense ID
SWIFT
IBAN
P000815
…
…
…
HD-MM 815
…
…
…
…
P004711
…
…
…
HD-ML 4711
…
…
…
…
P012345
…
…
…
HD-OI 2345
…
…
…
…
…
…
…
…
…
…
…
…
…
Here are 7 keys (provided there are no joint accounts) Just 1 key is selected as Primary Key
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Basic Concepts
What is a Foreign-Key?
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Foreign-Keys
Foreign-Key = set of columns, which is a (primary) key in an(other) table •
The foreign-key can refer to its own table
•
It is not necessary to share the same name(s)
•
The foreign-key can contain only those values that occur as a (primary) key value in another table (in addition and if applicable, NULL values are allowed)
•
The foreign-key is usually not a key!
Employee
©
DNumber D010000 D010001 D010002 D010003 D010004 D010005
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Name Mr A Ms B Mr C Ms D Mr E Ms F
DepartmentNR A01 A01 A02 A02 A02 A03
Foreign-key Relationship
Department DNr Function A01 HANA-Development A02 HANA-Sales A03 HANA-Training A04 ABAP-Development
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Foreign-Keys
A Foreign-Key can consist of multiple columns
Publication PaperID Topic P001 In-Memory P002 Column Store P003 Row Store P004 SQL & XML P005 XML & SQL P006 DB Recovery
Uni HPI HPI
StudentID 12345 12345
HPI FSU FSU FSU
77777 12345 12345 77777
Student
Uni StudentID Name FirstName HPI 12345 A B HPI 77777 C D FSU 12345 E F FSU 77777 G H
This example shows 1 Foreign-Key built on 2 columns
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Foreign-Keys
There can be multiple Foreign-Keys
Vendor
VID L1 L2 L3 …
©
Name Heidelberg Paper Wiesloch Paper Walldorf Paper …
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Delivery VID PNr Year Quantity
L1 L1 L1 L1 L2 L2 L3 …
A1 A1 A1 A2 A1 A2 A1 …
2010 2011 2012 2012 2011 2011 2012 …
320 570 925 102 577 100 999 …
Product PNr Description A1 Sticky Notes A2 Printing Paper A3 Envelopes … …
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Foreign-Keys
The foreign-key can refer to its own table
Employee
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DNumber D010000 D010001 D010002 D010003 D010004 D010005 D010006 D010007
Name Mr A Ms B Mr C Ms D Mr E Ms F Mr G Ms H
DNumberManager D010007 D010007 D010004 D010002 D010002 D010005 D010005
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Example Scenario
Fictional vehicle registration office as an example scenario
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Example: Registration Office
The database of a fictional registration office will serve as the basis for further explanations. The tables in this database have been specifically tailored to the SQL course and are not an example of good database design.
©
The officials working in the fictional registration office have a manager. Each vehicle is registered for exactly one owner (or is unregistered). There is a list of vehicles that have been reported stolen. Owners, who have at least three vehicles registered, are assigned to one or multiple contacts.
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Example: Registration Office
Official(PNr, Name, Overtime, Salary, Manager)
Contact(PersNumber, OwnerID) Owner(OwnerID, Name, Birthday, City)
Car(CarID, PlateNumber, Brand, Color, HP, Owner) Stolen(PlateNumber, reported_at)
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Officials
Official
©
PNr
Name
P01
Mr A
P02
Salary
Manager
10
A09
P04
Mr B
10
A10
P04
P03
Ms C
20
A09
P04
P04
Ms D
NULL
A12
P09
P05
Mr E
10
A08
P08
P06
Mr F
18
A09
P08
P07
Ms G
22
A11
P08
P08
Ms H
NULL
A13
P09
P09
Mr I
NULL
A14
NULL
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Overtime
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Owner
Owner
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OwnerID
Name
Birthday
City
H01
Ms T
20.06.1934
Wiesloch
H02
Ms U
11.05.1966
Hockenheim
H03
SAP AG
NULL
Walldorf
H04
HDM AG
NULL
Heidelberg
H05
Mr V
21.04.1952
Leimen
H06
Ms W
01.06.1957
Wiesloch
H07
IKEA
NULL
Walldorf
H08
Mr X
30.08.1986
Walldorf
H09
Ms Y
10.02.1986
Sinsheim
H10
Mr Z
03.02.1986
Ladenburg
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Contact
Contact
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PersNumber
OwnerID
P01
H03
P01
H04
P01
H07
P04
H03
P04
H04
P08
H04
P08
H07
P09
H03
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Cars Car
©
CarID
PlateNumber
Brand
Color
F01 F02 F03 F04 F05 F06 F07 F08 F09 F10
HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507
Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW
red black blue white black yellow blue black red black
75 120 184 136 170 260 116 160 105 140
H06 H03 H03 H07 H03 H05 H03 H07 H02 H04
F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 NULL HD-Y 333 HD-MQ 2006 HD-VW 2012 NULL
BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi
green red red white black green orange red black green
184 105 136 170 136 120 184 90 125 184
H02 H04 H07 H03 H03 NULL H09 H03 H01 NULL
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HP
Owner
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Stolen Cars Stolen
©
PlateNumber
reported_at
HD-VW 1999
20.06.2012
HD-V 106
01.06.2012
HD-Y 333
21.05.2012
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Registration Office & EU
A new (fictional) European Union directive requires that the information about which vehicle is registered to which owner, has to be stored in a central transnational database. The vehicle identification number (CarID) is unique across the EU, but not the Owner ID. Owner_EU(Country, OwnerID, Name, Birthday, City) Car_EU(CarID, PlateNumber, Brand, Color, HP, Country, Owner)
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Owner (EU-wide) Owner_EU
©
Country
OwnerID
Name
Birthday
City
D
H01
Ms T
20.06.1934
Wiesloch
D
H02
Ms U
11.05.1966
Hockenheim
D
H03
SAP AG
NULL
Walldorf
D
H04
HDM AG
NULL
Heidelberg
D
H05
Mr V
21.04.1952
Leimen
D
H06
Ms W
01.06.1957
Wiesloch
D
H07
IKEA
NULL
Walldorf
D
H08
Mr X
30.08.1986
Walldorf
D
H09
Ms Y
10.02.1986
Sinsheim
D
H10
Mr Z
03.02.1986
Ladenburg
A
H01
Ms O
21.05.1977
Wien
A
H02
Mr P
02.08.1977
Salzburg
E
H01
Señor Q
18.02.1925
Madrid
E
H02
Señora R
27.02.1927
Barcelona
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Cars (EU-wide) Car_EU
©
CarID F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 F21 F22 F23 F24
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PlateNumber HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 NULL HD-Y 333 HD-MQ 2006 HD-VW 2012 NULL W-302 ML S-215 MM 1409 EMM 3206 MLM
Brand Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi Mercedes VW Audi VW
Color red black blue white black yellow blue black red black green red red white black green orange red black green black silvern blue black
HP 75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184 170 184 116 170
Country D D D D D D D D D D D D D D D NULL D D D NULL A A E E
Owner H06 H03 H03 H07 H03 H05 H03 H07 H02 H04 H02 H04 H07 H03 H03 NULL H09 H03 H01 NULL H01 H02 H01 H02
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Unit 1: Motivation And Basic Concepts Summary
You should now be able to: • List some database models
• • • • •
Understand the motivation for and foundation of the relational model Understand why learning SQL is important when dealing with HANA Understand how the relation model and SQL are related List the three sub-languages of SQL Understand the sample database used throughout the course
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Internal
6969
Unit 2 Reading data from a table or view
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Unit 2: Reading Data From A Table Or View
After completing this unit, you will be able to: • Write simple database queries using SQL‘s SELECT statement • Project columns in and out of queries using the SELECT clause • Avoid duplicates in SELECT statement result sets • Include columns based on conditions • Use built-in functions in column lists and WHERE clauses • Limit results sets to the first N rows • Ensure a specific order in SELECT statement result sets • Restrict the result set using the WHERE clause
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Internal
7171
Reading Database Access
Which SQL statement can be used for reading data from a table or view?
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Reading Database Access
SELECT … FROM … WHERE … It
is used to read from a table or view
The SELECT statement is the central construct for read access to data (database) with SQL
The SELECT statement can include the following (optional) clauses:
WHERE, GROUP BY, HAVING, ORDER BY
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Reading Database Access
The SELECT statement
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SELECT Statement
SELECT FROM WHERE GROUP BY HAVING ORDER BY
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Column, Column, COUNT(*) Table Condition Column, Column Group_Condition Column ASC, Column DESC;
75
SELECT
SELECT clause
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76
SELECT Clause
You can specify a single column in the projection list:
SELECT Name FROM Official;
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
77
SELECT Clause
You can specify multiple columns in the projection list:
SELECT PNr, Name, Salary FROM Official;
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2013 SAP AG. All rights reserved.
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
Salary -----A09 A10 A09 A12 A08 A09 A11 A13 A14
78
SELECT Clause
The sequence of the columns in the projection list is relevant:
SELECT PNr, Name FROM Official; PNR --P01 P02 P03 P04 …
©
NAME ---Mr A Mr B Ms C Ms D …
2013 SAP AG. All rights reserved.
SELECT Name, PNr FROM Official; NAME ---Mr A Mr B Ms C Ms D …
PNR --P01 P02 P03 P04 …
79
SELECT Clause
The same column can appear several times in the projection list: We do not recommend using this option
SELECT PNr, PNr, PNr FROM Official;
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2013 SAP AG. All rights reserved.
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
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SELECT Clause
An asterisk (*) in the projection list represents "all columns“:
SELECT * FROM Official; PNR --P01 P02 P03 P04 …
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D …
OVERTIME -------10 10 20 ? …
SALARY -----A09 A10 A09 A12 …
MANAGER ------P04 P04 P04 P09 …
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SELECT Clause Columns can be specified in the projection list in addition to the asterisk (*): We do not recommend using this option
SELECT PNr, *, Name FROM Official; PNR --P01 P02 P03 P04 …
©
PNR --P01 P02 P03 P04 …
NAME ---Mr A Mr B Ms C Ms D …
2013 SAP AG. All rights reserved.
OVERTIME -------10 10 20 ? …
SALARY -----A09 A10 A09 A12 …
MANAGER ------P04 P04 P04 P09 …
NAME ---Mr A Mr B Ms C Ms D …
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SELECT Clause
The asterisk (*) can be used multiple times in the projection list: We do not recommend using this option
SELECT *, *, * FROM Official; PNR
NAME
OVERTIME
SALARY
MANAGER
PNR
NAME
OVERTIME
SALARY
MANAGER
PNR
NAME
OVERTIME
SALARY
MANAGER
---
----
--------
------
-------
---
----
--------
------
-------
---
----
--------
------
-------
P01
Mr A
10
A09
P04
P01
Mr A
10
A09
P04
P01
Mr A
10
A09
P04
P02
Mr B
10
A10
P04
P02
Mr B
10
A10
P04
P02
Mr B
10
A10
P04
P03
Ms C
20
A09
P04
P03
Ms C
20
A09
P04
P03
Ms C
20
A09
P04
P04
Ms D
?
A12
P09
P04
Ms D
?
A12
P09
P04
Ms D
?
A12
P09
P05
Mr E
10
A08
P08
P05
Mr E
10
A08
P08
P05
Mr E
10
A08
P08
P06
Mr F
18
A09
P08
P06
Mr F
18
A09
P08
P06
Mr F
18
A09
P08
P07
Ms G
22
A11
P08
P07
Ms G
22
A11
P08
P07
Ms G
22
A11
P08
P08
Ms H
?
A13
P09
P08
Ms H
?
A13
P09
P08
Ms H
?
A13
P09
P09
Mr I
?
A14
?
P09
Mr I
?
A14
?
P09
Mr I
?
A14
?
©
2013 SAP AG. All rights reserved.
83
SELECT Clause
You can generate additional “artificial” result columns:
SELECT 'Today', Name, 'has been assigned to salary group', Salary FROM Official; 'Today' ------Today Today Today Today Today …
©
NAME ---Mr A Mr B Ms C Ms D Mr E …
2013 SAP AG. All rights reserved.
'has been assigned to salary group' ----------------------------------has been assigned to salary group has been assigned to salary group has been assigned to salary group has been assigned to salary group has been assigned to salary group …
SALARY -----A09 A10 A09 A12 A08 …
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SELECT Clause
The additional artificial result column can have a numeric type:
SELECT 'The working week of', Name, 'amounts to', 40, 'hours.' FROM Official; 'The working week of' --------------------The working week of The working week of The working week of The working week of The working week of The working week of The working week of The working week of The working week of
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
'amounts to' -----------amounts to amounts to amounts to amounts to amounts to amounts to amounts to amounts to amounts to
40 -40 40 40 40 40 40 40 40 40
'hours'. -------hours. hours. hours. hours. hours. hours. hours. hours. hours.
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SELECT Clause
The projection list may only contain artificial result columns:
SELECT 'Good Morning!' FROM Official;
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2013 SAP AG. All rights reserved.
'Good Morning!' --------------Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning!
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SELECT Clause
If the projection list only contains artificial result columns, you can use the special table “Dummy” as reference:
SELECT 'Good Morning!' FROM Dummy;
'Good Morning!' --------------Good Morning!
The “DUMMY“ table contains one column and one row:
SELECT * FROM Dummy;
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2013 SAP AG. All rights reserved.
DUMMY ----X
87
SELECT Clause
SELECT a, 'b', "c", 1, '2', "3" FROM "4";
a
©
A … … …
'b' --b b b
c … … …
1 1 1 1
'2' --2 2 2
3 … … …
Existing column named “A“ (“A“ as capital letter)
'b'
Artificial result column with string “b“ as value in each row
"c"
Existing column named “c“ (“c“ as lower case letter)
1
Artificial result column with 1 as numeric value in each row
'2'
Artificial result column with string “2“ as value in each row
"3"
Existing column named “3“
"4"
Existing table named “4“
2013 SAP AG. All rights reserved.
88
SELECT Clause
The projection list can have computed columns. If a NULL value is used in an arithmetic operation, the result is also a NULL value:
SELECT Name, Overtime * 60 FROM Official;
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
OVERTIME*60 ----------600 600 1200 ? 600 1080 1320 ? ? 89
SELECT Clause
0 * NULL results in NULL (and not 0):
SELECT Name, Overtime, Overtime * 0 FROM Official; NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I ©
2013 SAP AG. All rights reserved.
OVERTIME -------10 10 20 ? 10 18 22 ? ?
OVERTIME*0 ---------0 0 0 ? 0 0 0 ? ? 90
SELECT Clause
You can explicitly name computed result columns:
SELECT Name, Overtime * 60 AS Overminutes FROM Official; NAME ---Mr A Mr B Ms C Ms D …
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2013 SAP AG. All rights reserved.
OVERMINUTES ----------600 600 1200 ? …
91
SELECT Clause
Quotation marks are required, if the result column name should include lowercase letters:
SELECT Name, Overtime * 60 AS "Overminutes" FROM Official; NAME ---Mr A Mr B Ms C Ms D …
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2013 SAP AG. All rights reserved.
Overminutes ----------600 600 1200 ? …
92
SELECT Clause
Quotation marks are also required, if the result column name should contain blanks:
SELECT Name, Overtime * 60 AS "overtime minutes" FROM Official; NAME ---Mr A Mr B Ms C Ms D …
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2013 SAP AG. All rights reserved.
overtime minutes ---------------600 600 1200 ? …
93
SELECT Clause
You can also rename non-calculated result columns:
SELECT PNr AS PersNumber, Salary AS "Salary Group" FROM Official; PERSNUMBER ---------P01 P02 P03 P04 …
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2013 SAP AG. All rights reserved.
Salary Group -----------A09 A10 A09 A12 …
94
SELECT Clause
In the (re)naming of result columns the keyword "AS“ can be omitted:
SELECT PNr PersNumber, Salary "Salary Group" FROM Official; PERSNUMBER ---------P01 P02 P03 P04 …
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2013 SAP AG. All rights reserved.
Salary Group -----------A09 A10 A09 A12 …
95
SELECT Clause
You can use existing column names for (re)naming of result columns:
SELECT PNr Salary, Salary PNr FROM Official;
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2013 SAP AG. All rights reserved.
SALARY -----P01 P02 P03 P04 P05 P06 P07 P08 P09
PNR --A09 A10 A09 A12 A08 A09 A11 A13 A14
96
SELECT Clause
Multiple result columns can have the same name: We do not recommend using this option
SELECT PNr AS xyz, Name AS xyz, Salary AS xyz FROM Official; XYZ --P01 P02 P03 P04 …
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2013 SAP AG. All rights reserved.
XYZ ---Mr A Mr B Ms C Ms D …
XYZ --A09 A10 A09 A12 …
97
SELECT Clause
You can use functions in the projection list. The corresponding result columns can be named explicitly.
Which owner is born in which year?
SELECT Name, YEAR(Birthday) AS "Year of Birth" FROM Owner;
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2013 SAP AG. All rights reserved.
NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z
Year of Birth ------------1934 1966 ? ? 1952 1957 ? 1986 1986 1986
98
SELECT Clause
When were vehicle owners first allowed to drive a car in Germany?
SELECT Name, ADD_YEARS(Birthday, 18) AS "18th Birthday" FROM Owner; NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z
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2013 SAP AG. All rights reserved.
18th Birthday ------------1952-06-20 1984-05-11 ? ? 1970-04-21 1975-06-01 ? 2004-08-30 2004-02-10 2004-02-03
99
SELECT Clause
Function calls can be nested On which day is the owner’s 18th birthday?
SELECT Name, DAYNAME ( ADD_YEARS (Birthday, ROUND(ABS(-18.2)) ) ) AS Weekday FROM Owner;
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2013 SAP AG. All rights reserved.
NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z
WEEKDAY ------FRIDAY FRIDAY ? ? TUESDAY SUNDAY ? MONDAY TUESDAY TUESDAY 100
Functions
Which functions are provided by SAP HANA?
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101
Functions
SAP HANA provides following functions (selection):
©
Function
Explanation
YEAR(Date)
year
ADD_YEARS(Date, n)
n years later
DAYNAME(Date)
weekday (English)
CURRENT_DATE
current date
ABS(Number)
absolute value
ROUND(Number)
rounding
SQRT(Number)
square root
UPPER(String)
convert to upper case
SUBSTR(String, Start, Length)
cut out of a string (substring)
LENGTH(String)
length of a string
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102
SELECT Clause
You can qualify the column name by adding the table name:
SELECT Official.Name FROM Official;
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
103
SELECT Clause
The qualification with the table name is also allowed, if the output table has to contain all columns:
SELECT Official.* FROM Official; PNR --P01 P02 P03 P04 …
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D …
OVERTIME -------10 10 20 ? …
SALARY -----A09 A10 A09 A12 …
MANAGER ------P04 P04 P04 P09 …
104
SELECT Clause
The projection list can include a mix of qualified and unqualified column names:
SELECT Name, Official.PNr FROM Official;
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NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
105
Tuple Variables Or Table Aliases
Using Tuple Variables, also known as Table Aliases
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106
Tuple Variables aka Table Aliases
You can use tuple variables in the projection list. Tuple variables are defined in the FROM clause and also referred to as table aliases in SAP HANA (and most other DBMSs).
SELECT o.Name, o.PNr FROM Official o; You can use the (optional) key word AS in the definition of a tuple variable:
SELECT o.Name, o.PNr FROM Official AS o;
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
107
Tuple Variables aka Table Aliases
Despite having defined table aliases the projection list may contain unqualified column names:
SELECT Name, PNr FROM Official o; A mix of qualified and unqualified column names is possible:
SELECT o.Name, PNr FROM Official o;
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2013 SAP AG. All rights reserved.
NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09
108
Tuple Variables aka Table Aliases
If a table alias is defined in the FROM clause, you are not allowed to use the corresponding table name for qualification of a column name:
SELECT Official.Name FROM Official o;
SELECT Official.Name FROM Official;
SELECT o.Name FROM Official o;
SELECT o.Name FROM Official;
SELECT Name FROM Official o;
SELECT Name FROM Official;
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109
Case Differentiation
Can the projection list contain columns that are based on a case differentiation?
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110
Case Differentiation
The projection list can contain columns that are based on a case differentiation These columns can be named explicitly:
SELECT *, CASE WHEN WHEN ELSE END AS FROM Car;
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HP < 120 THEN 'low' HP >= 120 AND HP < 180 THEN 'medium' 'high' Rating
111
Case Differentiation CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
©
PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 ? HD-Y 333 HD-MQ 2006 HD-VW 2012 ?
2013 SAP AG. All rights reserved.
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi
COLOR -----red black blue white black yellow blue black red black green red red white black green orange red black green
HP --75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184
OWNER ----H06 H03 H03 H07 H03 H05 H03 H07 H02 H04 H02 H04 H07 H03 H03 ? H09 H03 H01 ?
RATING -----low medium high medium medium high low medium low medium high low medium medium medium medium high low medium high
112
Case Differentiation
If a case differentiation does not correspond to any of the listed cases, a NULL value is returned:
SELECT CarID, CASE Color WHEN 'red' WHEN 'green' WHEN 'yellow' WHEN 'blue' WHEN 'white' WHEN 'black' END AS Color FROM Car;
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THEN THEN THEN THEN THEN THEN
'FF0000' '00FF00' 'FFFF00' '0000FF' 'FFFFFF' '000000'
CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
COLOR -----FF0000 000000 0000FF FFFFFF 000000 FFFF00 0000FF 000000 FF0000 000000 00FF00 FF0000 FF0000 FFFFFF 000000 00FF00 ? FF0000 000000 00FF00
113
Case Differentiation
Both the THEN and the ELSE branch can contain references to table columns:
SELECT CarID, CASE WHEN PlateNumber IS NOT NULL THEN PlateNumber ELSE 'The Car is not registered!' END AS PlateNumber, CASE Brand WHEN 'Mercedes' THEN 'Mercedes-Benz' WHEN 'VW' THEN 'Volkswagen' ELSE Brand END AS Brand, Color, HP FROM Car;
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114
Case Differentiation CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
©
PLATENUMBER -------------------------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 The Car is not registered! HD-Y 333 HD-MQ 2006 HD-VW 2012 The Car is not registered!
2013 SAP AG. All rights reserved.
BRAND ------------Fiat Volkswagen BMW Mercedes-Benz Mercedes-Benz Audi Audi Volkswagen Skoda BMW BMW Skoda Renault Mercedes-Benz Skoda Opel Audi Renault Volkswagen Audi
COLOR -----red black blue white black yellow blue black red black green red red white black green orange red black green
HP --75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184
115
Duplicate Elimination
SELECT DISTINCT
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116
Duplicate Elimination
A table with a primary key does not contain duplicates
SELECT * FROM Car; CARID ----… F04 F05 … F09 … F14 F15 …
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2013 SAP AG. All rights reserved.
PLATENUMBER ----------… HD-AL 1002 HD-MM 3206 … HD-UP 13 … HD-MM 1977 HD-MB 3030 …
BRAND -------… Mercedes Mercedes … Skoda … Mercedes Skoda …
COLOR ----… white black … red … white black …
HP --… 136 170 … 105 … 170 136 …
OWNER ----… H07 H03 … H02 … H03 H03 …
117
Duplicate Elimination
Duplicates can occur when a key column is not included in the projection list:
SELECT Brand FROM Car;
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2013 SAP AG. All rights reserved.
BRAND -------… Mercedes Mercedes … Skoda … Mercedes Skoda …
118
Duplicate Elimination
If duplicate elimination is not intended, you can specify the keyword ALL:
SELECT ALL Brand FROM Car;
©
2013 SAP AG. All rights reserved.
BRAND -------… Mercedes Mercedes … Skoda … Mercedes Skoda …
119
Duplicate Elimination
The keyword
DISTINCT ensures that the result table contains no duplicates:
SELECT DISTINCT Brand FROM Car;
©
2013 SAP AG. All rights reserved.
BRAND -------Fiat VW BMW Mercedes Audi Skoda Renault Opel
120
Duplicate Elimination
NULL values are treated in duplicate elimination as "normal" values The result table contains (at most) one row that consists entirely of NULL values:
SELECT DISTINCT Overtime FROM Official;
©
2013 SAP AG. All rights reserved.
OVERTIME -------10 20 ? 18 22
121
Duplicate Elimination The "duplicate" term always refers to full result rows:
SELECT Brand, Color FROM Car;
Duplicates
©
2013 SAP AG. All rights reserved.
BRAND -------… Mercedes Mercedes Skoda Skoda Mercedes Skoda …
COLOR ----… white black red red white black …
122
Duplicate Elimination
If a projection list contains multiple columns, DISTINCT always refers to the combination of all these columns:
SELECT DISTINCT Brand, Color FROM Car;
©
2013 SAP AG. All rights reserved.
BRAND -------… Mercedes Mercedes Skoda Skoda …
COLOR ----… white black red black …
123
Duplicate Elimination
DISTINCT can also be used if the result table should contain all the columns If the (source) table has a (primary) key DISTINCT is not required: SELECT DISTINCT * FROM Car; CARID ----… F04 F05 … F09 … F14 F15 …
©
2013 SAP AG. All rights reserved.
PLATENUMBER ----------… HD-AL 1002 HD-MM 3206 … HD-UP 13 … HD-MM 1977 HD-MB 3030 …
BRAND -------… Mercedes Mercedes … Skoda … Mercedes Skoda …
COLOR ----… white black … red … white black …
HP --… 136 170 … 105 … 170 136 …
OWNER ----… H07 H03 … H02 … H03 H03 …
124
Sorting
ORDER BY
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2013 SAP AG. All rights reserved.
125
Sorting
The result table can be sorted by a specific column:
SELECT Brand, Color FROM Car ORDER BY Brand;
©
2013 SAP AG. All rights reserved.
BRAND -------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW
COLOR -----yellow blue orange green blue black green red white black white green red red red red black black black black 126
Sorting
A descending sorting is possible:
SELECT Brand, Color FROM Car ORDER BY Brand DESC;
©
2013 SAP AG. All rights reserved.
BRAND -------VW VW VW Skoda Skoda Skoda Renault Renault Opel Mercedes Mercedes Mercedes Fiat BMW BMW BMW Audi Audi Audi Audi
COLOR -----black black black black red red red red green white black white red green black blue green orange blue yellow 127
Sorting
To sort ascending apply the optional keyword
SELECT Brand, Color FROM Car ORDER BY Brand ASC;
©
2013 SAP AG. All rights reserved.
ASC:
BRAND -------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW
COLOR -----yellow blue orange green blue black green red white black white green red red red red black black black black
128
Sorting
The sorting can be applied using a column that does not appear in the projection list:
SELECT Brand, Color FROM Car ORDER BY PlateNumber;
©
2013 SAP AG. All rights reserved.
BRAND -------Opel Audi Renault Mercedes VW BMW Skoda Audi Mercedes BMW Mercedes Renault BMW Skoda Fiat Audi VW VW Skoda Audi
COLOR -----green green red white black blue black blue white green black red black red red yellow black black red orange 129
Sorting
You can sort using a combination of columns:
SELECT Brand, Color FROM Car ORDER BY Brand ASC, Color DESC;
©
2013 SAP AG. All rights reserved.
BRAND -------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW
COLOR -----yellow orange green blue green blue black red white white black green red red red red black black black black 130
Sorting
Instead of the column names in the ORDER BY clause, the column numbers (based on the projection list) can be used:
SELECT Brand, Color FROM Car ORDER BY 1 ASC, 2 DESC;
©
2013 SAP AG. All rights reserved.
BRAND -------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW
COLOR -----yellow orange green blue green blue black red white white black green red red red red black black black black 131
Sorting
If result columns are named explicitly, you can refer to the new name for sorting:
SELECT Brand AS Manufacturer, Color FROM Car ORDER BY Manufacturer ASC, Color DESC;
©
2013 SAP AG. All rights reserved.
MANUFACTURER -----------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW
COLOR -----yellow orange green blue green blue black red white white black green red red red red black black black black 132
Sorting
If result columns are explicitly renamed, you can still reference the original name in the ORDER BY clause:
SELECT Brand AS Manufacturer, Color FROM Car ORDER BY Brand ASC, Color DESC;
©
2013 SAP AG. All rights reserved.
MANUFACTURER -----------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW
COLOR -----yellow orange green blue green blue black red white white black green red red red red black black black black 133
Sorting
You can sort based on calculated values. Sorting criteria: How much is the power below 200 kW?
SELECT CarID, Brand, HP FROM Car ORDER BY 200 - HP / 1.36 ASC;
©
2013 SAP AG. All rights reserved.
CARID ----F06 F03 F11 F17 F20 F05 F14 F08 F10 F04 F13 F15 F19 F02 F16 F07 F09 F12 F18 F01
BRAND -------Audi BMW BMW Audi Audi Mercedes Mercedes VW BMW Mercedes Renault Skoda VW VW Opel Audi Skoda Skoda Renault Fiat
HP --260 184 184 184 184 170 170 160 140 136 136 136 125 120 120 116 105 105 90 75 134
Sorting
You can reference functions in the ORDER BY clause:
SELECT Name, Birthday FROM Owner ORDER BY YEAR(Birthday) DESC, Name ASC; NAME -----Mr X Mr Z Ms Y Ms U Ms W Mr V Ms T HDM AG IKEA SAP AG ©
2013 SAP AG. All rights reserved.
BIRTHDAY ---------1986-08-30 1986-02-03 1986-02-10 1966-05-11 1957-06-01 1952-04-21 1934-06-20 ? ? ? 135
Top-n Clause
How many rows are in the result set?
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2013 SAP AG. All rights reserved.
136
Top-n Clause
You can determine how many rows should be included in the query result (maximum). What are the ten most powerful vehicles (based on horse power)?
SELECT TOP 10 CarID, Brand, Color, HP FROM Car CARID BRAND ORDER BY 4 DESC; ----- -------F06 F20 F17 F11 F03 F14 F05 F08 F10 F15 ©
2013 SAP AG. All rights reserved.
Audi Audi Audi BMW BMW Mercedes Mercedes VW BMW Skoda
COLOR -----yellow green orange green blue white black black black black
HP --260 184 184 184 184 170 170 160 140 136 137
Top-n Clause
You can also use the Top-n Clause if the result table should include all columns. What are the 5 most powerful vehicles (based on horse power)?
SELECT TOP 5 * FROM Car ORDER BY HP DESC; CARID ----F06 F20 F17 F11 F03
©
2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-VW 1999 ? HD-Y 333 HD-MM 208 HD-JA 1972
BRAND ----Audi Audi Audi BMW BMW
COLOR -----yellow green orange green blue
HP --260 184 184 184 184
OWNER ----H05 ? H09 H02 H03
138
Top-n Clause
You can combine the Top-n Clause with the keyword DISTINCT. What are the 7 highest HP values?
SELECT TOP 7 DISTINCT HP FROM Car ORDER BY 1 DESC;
©
2013 SAP AG. All rights reserved.
HP --260 184 170 160 140 136 125
139
Top-n Clause
No error is thrown if you request more rows than available. The result set will not be filled with additional, “artificial” rows. 570 different colors are requested:
SELECT TOP 570 DISTINCT Color FROM Car;
©
2013 SAP AG. All rights reserved.
COLOR -----red black blue white yellow green orange
140
Top-n Clause
It is possible to request 0 result rows. In this case the result set does not contain any row. Zero colors are requested:
SELECT TOP 0 Color FROM Car;
©
2013 SAP AG. All rights reserved.
COLOR -----
141
Limit And Offset Clause
You can use the Limit clause as alternative to the Top-n clause. The Limit clause comes at the very end of the SELECT statement. What are the 5 most powerful vehicles (based on horse power)?
SELECT * FROM Car ORDER BY HP DESC LIMIT 5; CARID ----F06 F20 F17 F11 F03
©
2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-VW 1999 ? HD-Y 333 HD-MM 208 HD-JA 1972
BRAND ----Audi Audi Audi BMW BMW
COLOR -----yellow green orange green blue
HP --260 184 184 184 184
OWNER ----H05 ? H09 H02 H03
142
Limit And Offset Clause
The Limit clause can be combined with the Offset clause to skip records. This allows you to read result sets “page by page”. What are the next 5 most powerful vehicles (based on horse power)?
SELECT * FROM Car ORDER BY HP DESC LIMIT 5 CARID OFFSET 5; ----F78 F77 F14 F05 F07
©
2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-MT 2510 HD-MT 2509 HD-MM 1977 HD-MM 3206 HD-IK 1002
BRAND -------? ? Mercedes Mercedes VW
COLOR -----green red white black black
HP --260 184 184 184 184
OWNER ----? ? H03 H03 H07
143
WHERE Clause
Which rows are included in the result set?
©
2013 SAP AG. All rights reserved.
144
WHERE Clause
The WHERE clause is used to filter rows. It is used to extract only those rows that fulfill a specified criterion.
SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'BMW'; PLATENUMBER ----------HD-JA 1972 HD-MT 507 HD-MM 208
©
2013 SAP AG. All rights reserved.
BRAND ----BMW BMW BMW
COLOR ----blue black green
145
WHERE Clause
You can reference numeric columns in the WHERE clause.
SELECT HP, Brand, Color FROM Car WHERE HP >= 170;
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HP --184 170 260 184 170 184 184
BRAND -------BMW Mercedes Audi BMW Mercedes Audi Audi
COLOR -----blue black yellow green white orange green
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WHERE Clause
You can reference a column in the WHERE clause that is not included in the projection list.
SELECT Brand, Color FROM Car WHERE HP >= 170;
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2013 SAP AG. All rights reserved.
BRAND -------BMW Mercedes Audi BMW Mercedes Audi Audi
COLOR -----blue black yellow green white orange green
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WHERE Clause
You can check for NULL in the WHERE clause:
SELECT CarID, Brand, Color FROM Car WHERE PlateNumber IS NULL; CARID ----F16 F20
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BRAND ----Opel Audi
COLOR ----green green
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WHERE Clause
You can check for IS in the WHERE clause:
NOT NULL
SELECT CarID, Brand, Color FROM Car WHERE PlateNumber IS NOT NULL;
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CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F17 F18 F19
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Audi Renault VW
COLOR -----red black blue white black yellow blue black red black green red red white black orange red black
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WHERE Clause
You can check if values are included IN a value-list.
SELECT Brand, Color FROM Car WHERE Color IN ('red', 'blue', 'orange'); BRAND ------Fiat BMW Audi Skoda Skoda Renault Audi Renault ©
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COLOR -----red blue blue red red red orange red 150
WHERE Clause
You can check if values are included in an interval.
SELECT PlateNumber, Brand, Color, HP FROM Car WHERE HP BETWEEN 140 AND 170; PLATENUMBER ----------HD-MM 3206 HD-IK 1002 HD-MT 507 HD-MM 1977
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BRAND -------Mercedes VW BMW Mercedes
COLOR ----black black black white
HP --170 160 140 170
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WHERE Clause
You can reference calculated values in the WHERE clause. We do not recommend this option for performance reasons. Which cars have a power of at least 125 kW?
SELECT CarID, Brand, HP FROM Car WHERE HP / 1.36 >= 125;
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CARID ----F03 F05 F06 F11 F14 F17 F20
BRAND -------BMW Mercedes Audi BMW Mercedes Audi Audi
HP --184 170 260 184 170 184 184
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WHERE Clause
You can use functions in the WHERE clause.
SELECT * FROM Owner WHERE YEAR(Birthday) = 1986; OWNERID ------H08 H09 H10
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NAME ---Mr X Ms Y Mr Z
BIRTHDAY ---------1986-08-30 1986-02-10 1986-02-03
CITY --------Walldorf Sinsheim Ladenburg
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WHERE Clause
You can use search patterns in the WHERE clause.
SELECT PlateNumber, Brand, Color, HP FROM Car WHERE PlateNumber LIKE '%MM%'; PLATENUMBER ----------HD-MM 3206 HD-MM 208 HD-MM 1977
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BRAND -------Mercedes BMW Mercedes
COLOR ----black green white
HP --170 184 170
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LIKE Predicate The wildcard character % (percentage sign) represents any string containing no, one, or multiple characters. The wildcard character _ (underscore) represents any single character. WHERE MyColumn LIKE 'M%'
String starting with “M“
WHERE MyColumn LIKE 'M _'
String of two characters starting with “M“
WHERE MyColumn LIKE '%M'
String ending with “M“
WHERE MyColumn LIKE '%M%'
String containing “M“
WHERE MyColumn LIKE '_ _ _'
String with length 3
WHERE MyColumn LIKE '_ _ _ _T_M%'
“T“ on fifth and “M“ on seventh position
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LIKE Predicate If you want to search for the percentage sign (%) or underscore (_) itself, you have to place an ESCAPE character in front. You can choose the ESCAPE character (with some restrictions). LIKE '$%%' ESCAPE '$'
String starting with a percentage sign
LIKE '$_ _' ESCAPE '$'
String of two characters starting with an underscore
LIKE '%$%' ESCAPE '$'
String ending with a percentage sign
LIKE '%$%%' ESCAPE '$'
String containing a percentage sign
LIKE '%$_$%%' ESCAPE '$'
String containing an underscore followed by a percentage sign
LIKE '_ _ _ _ $%_ $ _%' ESCAPE '$'
“%“ on fifth and “_“ on seventh position.
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LIKE Predicate If you need to search for the escape character within a string, you need to put the escape character in front of itself to mask it. LIKE '$%%$$' ESCAPE '$'
String starting with a percentage sign and ending with $
LIKE '$_%$$%$_' ESCAPE '$'
String starting and ending with an underscore and containing the escape character $
LIKE '%$%$$%' ESCAPE '$'
String containing a percentage sign followed by $
LIKE '%$_%$$' ESCAPE '$'
String containing an underscore and ending with $
LIKE '_ _ _ _ $%_ $$%' ESCAPE '$'
“%“ on fifth and “$“ on seventh position
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WHERE Clause
You can use compound WHERE clauses.
SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'Skoda' AND Color = 'red'; PLATENUMBER ----------HD-UP 13 HD-XY 4711
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BRAND ----Skoda Skoda
COLOR ----red red
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WHERE Clause
You can reference the same column multiple times.
SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'Skoda' OR Brand = 'BMW'; PLATENUMBER ----------HD-JA 1972 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-MB 3030
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BRAND ----BMW Skoda BMW BMW Skoda Skoda
COLOR ----blue red black green red black
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WHERE Clause
You can use brackets.
SELECT PlateNumber, Brand, Color FROM Car WHERE (Brand = 'Skoda' OR Brand = 'BMW') AND Color = 'black'; PLATENUMBER ----------HD-MT 507 HD-MB 3030
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BRAND ----BMW Skoda
COLOR ----black black
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WHERE Clause The license plate started with "HD" for Heidelberg I'm sure I saw an "M" I clearly remember the digits "2" and "6" The "2" was definitely left of the "6" Between the "2" and the "6" was exactly one digit It was neither Skoda, nor a VW The car was blue or green CARID ----F07
PLATENUMBER ----------HD-ML 3206
BRAND ----Audi
COLOR ----blue
HP --116
SELECT * FROM Car WHERE PlateNumber LIKE 'HD-%M% %2_6%' AND Brand 'Skoda' AND Brand 'VW' AND (Color = 'blue' OR Color = 'green');
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OWNER ----H03
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Operator Precedence
What is the precedence of the different operators?
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Operator Precedence
The operators have the precedence indicated by the table below: Precedence
Operator
Explanation
Highest
()
parentheses
‒
unary minus
*, /
multiplication, division
+, ‒
addition, subtraction
=, =, , IS NULL, LIKE, BETWEEN
Lowest
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comparison
NOT
logical negation
AND
conjunction
OR
disjunction
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Unit 2: Reading Data From A Table And View Summary
You should now be able to: • Write simple database queries using SQL‘s SELECT statement
• • • • • • •
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Project columns in and out of queries using the SELECT clause Avoid duplicates in SELECT statement result sets Include columns based on conditions Use built-in functions in column lists and WHERE clauses Limit results sets to the first N rows Ensure a specific order in SELECT statement result sets Restrict the result set using the WHERE clause
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164
Unit 3 Aggregating Data
165
Unit 3: Aggregating Data Learning Objectives After completing this unit, you will be able to: • Determine aggregated values on table columns using a single SELECT statement • List the aggregate functions supported by HANA • Determine such aggregated values for groups of rows, using the GROUP BY clause • Filter such groups using the HAVING clause
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Aggregate Expressions
Calculations across multiple rows
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Aggregate Expressions
Count(*)
You can calculate the number of rows in the result set using COUNT .
Rows containing only NULL values are included.
What is the quantity of cars with the brand “Audi“?
SELECT COUNT(*) FROM Car WHERE Brand = 'Audi';
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COUNT(*) -------4
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Count(*)
Aggregate Expressions
You can explicitly name columns created by aggregate expressions.
What is the quantity of cars with the brand “Audi“?
SELECT COUNT(*) AS "Quantity of Audi" FROM Car WHERE Brand = 'Audi'; Quantity of Audi ---------------4
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Aggregate Expressions
Count()
You can calculate the number of values within a single column. NULL values are not included.
You can only use a single column as parameter of COUNT.
How many cars are registered to an owner? This does not calculate the number of different owners!
SELECT COUNT(Owner) FROM Car;
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2013 SAP AG. All rights reserved.
COUNT(OWNER) -----------18
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Aggregate Expressions
Count(DISTINCT )
You can calculate the number of different NOT-NULL-values of a certain column. You can use only a single column as parameter for COUNT DISTINCT. NULL values are not included.
How many different owners have a registered car?
SELECT COUNT(DISTINCT Owner) FROM Car; COUNT(DISTINCT OWNER) --------------------8
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Aggregate Expressions
SELECT * FROM T; SELECT S FROM T;
S ? ? ? X X
SELECT COUNT(*) FROM T;
COUNT(*) -------5
SELECT COUNT(S) FROM T;
COUNT(S) -------2
SELECT COUNT(DISTINCT *) FROM T; SELECT COUNT(DISTINCT S) FROM T;
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COUNT(DISTINCT S) ----------------1
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Aggregate Expressions
MIN/MAX()
You can calculate the minimum or maximum value in a column.
What is the horsepower range of the registered cars?
SELECT MIN(HP), MAX(HP) FROM Car;
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MIN(HP) ------75
MAX(HP) ------260
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Aggregate Expressions
You can combine aggregate expressions and “normal” functions.
In which year was the youngest owner born?
SELECT MAX(YEAR(Birthday)) AS Year FROM Owner;
YEAR ---1986
SELECT YEAR(MAX(Birthday)) AS Year FROM Owner;
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Aggregate Expressions
The sequence of nested functions can be relevant.
What is the lowest or highest HP value?
SELECT ABS(MAX(0 - HP)) FROM Car;
SELECT MAX(ABS(0 - HP)) FROM Car;
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ABS(MAX(0-HP)) -------------75
MAX(ABS(0-HP)) -------------260
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Aggregate Expressions
The WHERE clause is included in the minimum and maximum calculation.
What is the horsepower range of BMWs?
SELECT MIN(HP), MAX(HP) FROM Car WHERE Brand = 'BMW';
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MIN(HP) ------140
MAX(HP) ------184
176
Aggregate Expressions
SUM()
You can sum up the values in a column. The WHERE clause is included in the summation. NULL values in the column are ignored. Individual NULL values contained in the column do not result in NULL for the sum. (as long as there is at least one numeric value)
What is the sum of horsepower for all Audis?
SELECT SUM(HP) FROM Car WHERE Brand = 'Audi';
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SUM(HP) ------744
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Aggregate Expressions
You can sum up the distinct values of a column. The WHERE clause is included in the summation. NULL values of the column are ignored. Individual NULL values contained in the column do not result in NULL for the sum. (as long as there is at least one numeric value) If a value occurs multiple times, it is still counted only once when summing
SELECT SUM(DISTINCT HP) FROM Car WHERE Brand = 'Audi';
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SUM(DISTINCT )
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SUM(DISTINCT HP) ---------------560
178
Aggregate Expressions
AVG()
You can calculate the average of all values in a column. The WHERE clause is included in the calculation. NULL values in the column are completely ignored (in the numerator and denominator) Individual NULL values contained in the column do not result in NULL for the average.
What is the average horsepower of Audi?
SELECT AVG(HP) FROM Car WHERE Brand = 'Audi';
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AVG(HP) ------186
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Aggregate Expressions
AVG(DISTINCT )
You can calculate the average of all distinct values in a column. The WHERE clause is included in the calculation. NULL values in the column are completely ignored (in the numerator and denominator) Individual NULL values contained in the column do not result in NULL for the average. If a value occurs multiple times, it is still counted only once for the average calculation.
SELECT AVG(DISTINCT HP) FROM Car WHERE Brand = 'Audi'; AVG(DISTINCT HP) ----------------------------------186.6666666666666666666666666666667
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Aggregate Expressions
Which aggregate expressions are available in SAP HANA?
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181
Aggregate Expressions
SAP HANA provides the following aggregate expressions:
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Aggregate Name
Description
COUNT
Count
MIN
Minimum
MAX
Maximum
SUM
Sum
AVG
Average
STDDEV
Standard Deviation
VAR
Variance
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Group By
GROUP BY
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183
Group By
A table can be divided into (disjoint) groups of rows A group is represented in the query result by a single row Aggregate expressions will be evaluated separately for each group
What is the number of cars per brand?
SELECT Brand, COUNT(*) FROM Car GROUP BY Brand;
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BRAND -------Fiat VW BMW Mercedes Audi Skoda Renault Opel
COUNT(*) -------1 3 3 3 4 3 2 1
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Group By
What is the highest horse power value per brand?
SELECT Brand, MAX(HP) FROM Car GROUP BY Brand;
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2013 SAP AG. All rights reserved.
BRAND -------Fiat VW BMW Mercedes Audi Skoda Renault Opel
MAX(HP) ------75 160 184 170 260 136 136 120
185
Group By
NULL values of the GROUP BY column are treated as “normal” values creating a single group. How often a certain overtime value occurs?
SELECT Overtime, COUNT(*) AS Frequency FROM Official GROUP BY Overtime; OVERTIME FREQUENCY -------10 20 ? 18 22
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--------3 1 3 1 1
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Group By
You can combine GROUP BY with ORDER BY for sorting.
SELECT FROM GROUP BY ORDER BY
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Brand, MAX(HP) Car Brand 2 DESC, 1 ASC;
BRAND -------Audi BMW Mercedes VW Renault Skoda Opel Fiat
MAX(HP) ------260 184 170 160 136 136 120 75
187
Group By
You can use functions in the GROUP BY clause.
What is the number of owners per year of birth?
SELECT FROM GROUP BY ORDER BY
YEAR(Birthday), COUNT(*) Owner YEAR(Birthday) 2 DESC, 1 ASC; YEAR(BIRTHDAY) -------------? 1986 1934 1952 1957 1966
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COUNT(*) -------3 3 1 1 1 1
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Group By
The WHERE clause is processed before the grouping. What is the number of black cars per brand? Only brands with black cars are included into the result set.
SELECT FROM WHERE GROUP BY
©
Brand, COUNT(*) Car Color = 'black' Brand;
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BRAND -------VW Mercedes BMW Skoda
COUNT(*) -------3 1 1 1
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Group By
You can use a combination of columns in the GROUP BY clause.
SELECT Brand, Color, COUNT(*) FROM Car GROUP BY Brand, Color;
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2013 SAP AG. All rights reserved.
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi Skoda BMW BMW Renault Skoda Opel Audi Audi
COLOR -----red black blue white black yellow blue red black green red black green orange green
COUNT(*) -------1 3 1 2 1 1 1 2 1 1 2 1 1 1 1
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Group By
You can explicitly rename result columns in combination with GROUP BY.
SELECT Brand AS Manufacturer, Color, COUNT(*) AS "# of cars" FROM Car GROUP BY Brand, Color; MANUFACTURER -----------Fiat VW BMW Mercedes Mercedes Audi Audi Skoda BMW BMW Renault …
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COLOR -----red black blue white black yellow blue red black green red …
# of cars --------1 3 1 2 1 1 1 2 1 1 2 …
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HAVING Clause
What groups are included in the query result?
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192
HAVING
Using the HAVING clause you can specify which conditions a group must meet to be included in the result set. Which combinations of brand and color occur at least twice?
SELECT FROM GROUP BY HAVING
Brand, Color, COUNT(*) Car Brand, Color COUNT(*) >= 2; BRAND -------VW Mercedes Skoda Renault
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COLOR ----black white red red
COUNT(*) -------3 2 2 2
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HAVING
The HAVING condition can reference columns not included in the projection list. What is the number of brand-color combinations, where at least one car has less than 120 HP? We want to analyze based on the quantity of all cars, not only the cars with less than 120 HP.
SELECT FROM GROUP BY HAVING
Brand, Color, COUNT(*) Car Brand, Color MIN(HP) < 120; BRAND ------Fiat Audi Skoda Renault
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2013 SAP AG. All rights reserved.
COLOR ----red blue red red
COUNT(*) -------1 1 2 2
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SELECT statement
What is the number of cars per brand, which are black or red? Rename the Brand column to "Manufacturer" Display only those manufacturers with at least 2 cars Sort the result set first descending by number of cars Sort the result set second ascending by manufacturer.
SELECT FROM WHERE GROUP BY HAVING ORDER BY
©
c.Brand AS "Manufacturer", COUNT(*) Car c Color IN ('black', 'red') Brand COUNT(*) >= 2 Manufacturer 2 DESC, Brand ASC;
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-----------Skoda VW Renault
COUNT(*) -------3 3 2
195
Unit 3: Aggregating Data Summary
You should now be able to: • Determine aggregated values on table columns using a single SELECT statement • List the aggregate functions supported by HANA • Determine such aggregated values for groups of rows, using the GROUP BY clause • Filter such groups using the HAVING clause
©
2013 SAP AG. All rights reserved.
196
Unit 4 Reading Data From Multiple Tables Part I
197
Unit 4: Reading Data From Multiple Tables Part I Learning Objectives After completing this unit, you will be able to: • Merge the result of several select statements using the UNION statement • • •
©
Combine data from several tables when querying data using JOIN constructs List the various types of Joins Explain the differences between the various types of Joins, and apply the right type of Join depending on the problem
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198
Access to multiple tables
How can I read data from multiple tables and views?
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199
Access to multiple tables
For read access to multiple database tables / views the following options are available: Combination of results from several partial queries (UNION) UNION ALL vs. UNION
Combination of several tables (JOIN) CROSS JOIN vs. INNER JOIN vs. OUTER JOIN Implicit JOIN vs. Explicit JOIN
Nested queries (Sub Query) Uncorrelated Sub Query vs. Correlated Sub Query
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200
UNION [ALL]
Combination of results from several partial queries
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201
UNION [ALL]
SELECT Column, Column, Column FROM Table WHERE Condition UNION ALL SELECT Column, Column, Column FROM Table WHERE Condition;
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UNION [ALL]
You can combine the result tables of multiple queries using UNION [ALL].
The individual result tables must have the same number of columns. The corresponding result columns must have compatible data types. The column names of the resulting output table are based on the first SELECT statement.
SELECT PNr, Name FROM Official WHERE Salary = 'A09' UNION ALL SELECT OwnerID, Name FROM Owner WHERE Birthday >= '1977-05-21';
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PNR --P01 P03 P06 H08 H09 H10
NAME ---Mr A Ms C Mr F Mr X Ms Y Mr Z
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UNION [ALL]
Almost the same statement as on the previous slide – but with a changed sequence of partial queries. The column names of the resulting output table are based on the first SELECT statement.
SELECT OwnerID, Name FROM Owner WHERE Birthday >= '1977-05-21' UNION ALL SELECT PNr, Name FROM Official WHERE Salary = 'A09';
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2013 SAP AG. All rights reserved.
OWNERID ------H08 H09 H10 P01 P03 P06
NAME ---Mr X Ms Y Mr Z Mr A Ms C Mr F
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UNION [ALL]
With UNION [ALL] you can also explicitly rename the result columns. SELECT OwnerID AS "Person ID", Name FROM Owner WHERE Birthday >= '1977-05-21' UNION ALL SELECT PNr, Name FROM Official WHERE Salary = 'A09';
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2013 SAP AG. All rights reserved.
Person ID --------H08 H09 H10 P01 P03 P06
NAME ---Mr X Ms Y Mr Z Mr A Ms C Mr F
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UNION ALL
If the results of multiple partial queries overlap, the overall result includes duplicates with UNION ALL. SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'Mercedes' UNION ALL SELECT PlateNumber, Brand, Color FROM Car WHERE Color = 'white';
Duplicates
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2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-AL 1002 HD-MM 3206 HD-MM 1977 HD-AL 1002 HD-MM 1977
BRAND -------Mercedes Mercedes Mercedes Mercedes Mercedes
COLOR ----white black white white white
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UNION
You use UNION instead of UNION ALL to eliminate duplicates.
SELECT FROM WHERE UNION SELECT FROM WHERE
PlateNumber, Brand, Color Car Brand = 'Mercedes' PlateNumber, Brand, Color Car Color = 'white'; PLATENUMBER ----------HD-AL 1002 HD-MM 3206 HD-MM 1977
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BRAND -------Mercedes Mercedes Mercedes
COLOR ----white black white 207
UNION [ALL]
You can use UNION [ALL] to combine result tables of multiple partial queries. SELECT FROM WHERE UNION SELECT FROM WHERE UNION SELECT FROM WHERE
©
PlateNumber, Brand, Color Car Brand = 'BMW' PlateNumber, Brand, Color Car Color = 'yellow'
PlateNumber, Brand, Color Car Color = 'orange';
2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-JA 1972 HD-MT 507 HD-MM 208 HD-VW 1999 HD-Y 333
BRAND ----BMW BMW BMW Audi Audi
COLOR -----blue black green yellow orange 208
Joining Tables
Joining tables
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209
Joining Tables
With JOIN semantics you can distinguish between: CROSS JOIN INNER JOIN OUTER JOIN –
LEFT OUTER JOIN
–
RIGHT OUTER JOIN
–
FULL OUTER JOIN
With JOIN syntax you can distinguish between: Implicit JOIN Explicit JOIN
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210
Joining Tables
CROSS JOIN (Cartesian Product)
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211
CROSS JOIN
Each row of the left table is connected to each row of the right table. There is no CROSS JOIN join condition.
L
©
… … … … … …
X 1 2 3 4 5
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… … … … … …
… … … … … …
R
… … … … … …
Y 3 4 5 6 7
… … … … … …
… … … … … …
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CROSS JOIN
SELECT Column, Column, Column FROM Table, Table WHERE Condition; SELECT Column, Column, Column FROM Table CROSS JOIN Table WHERE Condition;
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Implicit CROSS JOIN
Each row of a table is connected to each row of the other table. The result set contains 10 * 20 = 200 rows
SELECT * FROM Owner, Car; OWNERID ------H01 H02 H03 … H01 H02 H03 … H01 H02 H03 … H10
©
2013 SAP AG. All rights reserved.
NAME -----Ms T Ms U SAP AG … Ms T Ms U SAP AG … Ms T Ms U SAP AG … Mr Z
BIRTHDAY ---------1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1986-02-03
CITY ---------Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Ladenburg
CARID ----F01 F01 F01 … F02 F02 F02 … F03 F03 F03 … F20
PLATENUMBER ----------HD-V 106 HD-V 106 HD-V 106 … HD-VW 4711 HD-VW 4711 HD-VW 4711 … HD-JA 1972 HD-JA 1972 HD-JA 1972 … ?
BRAND ----Fiat Fiat Fiat … VW VW VW … BMW BMW BMW … Audi
COLOR ----red red red … black black black … blue blue blue … green
HP --75 75 75 … 120 120 120 … 184 184 184 … 184
OWNER ----H06 H06 H06 … H03 H03 H03 … H03 H03 H03 … ?
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Explicit CROSS JOIN
Each row of a table is connected to each row of the other table. The result set contains 10 * 20 = 200 rows
SELECT * FROM Owner CROSS JOIN Car; OWNERID ------H01 H02 H03 … H01 H02 H03 … H01 H02 H03 … H10
©
2013 SAP AG. All rights reserved.
NAME -----Ms T Ms U SAP AG … Ms T Ms U SAP AG … Ms T Ms U SAP AG … Mr Z
BIRTHDAY ---------1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1986-02-03
CITY ---------Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Ladenburg
CARID ----F01 F01 F01 … F02 F02 F02 … F03 F03 F03 … F20
PLATENUMBER ----------HD-V 106 HD-V 106 HD-V 106 … HD-VW 4711 HD-VW 4711 HD-VW 4711 … HD-JA 1972 HD-JA 1972 HD-JA 1972 … ?
BRAND ----Fiat Fiat Fiat … VW VW VW … BMW BMW BMW … Audi
COLOR ----red red red … black black black … blue blue blue … green
HP --75 75 75 … 120 120 120 … 184 184 184 … 184
OWNER ----H06 H06 H06 … H03 H03 H03 … H03 H03 H03 … ?
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CROSS JOIN
You can specify a WHERE clause for the implicit and explicit CROSS JOIN.
SELECT * FROM Owner, Car WHERE HP > 250; OWNERID ------H01 H02 H03 H04 H05 H06 H07 H08 H09 H10
©
NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z
2013 SAP AG. All rights reserved.
BIRTHDAY ---------1934-06-20 1966-05-11 ? ? 1952-04-21 1957-06-01 ? 1986-08-30 1986-02-10 1986-02-03
SELECT * FROM Owner CROSS JOIN Car WHERE HP > 250; CITY ---------Wiesloch Hockenheim Walldorf Heidelberg Leimen Wiesloch Walldorf Walldorf Sinsheim Ladenburg
CARID ----F06 F06 F06 F06 F06 F06 F06 F06 F06 F06
PLATENUMBER ----------HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999
BRAND ----Audi Audi Audi Audi Audi Audi Audi Audi Audi Audi
COLOR -----yellow yellow yellow yellow yellow yellow yellow yellow yellow yellow
HP --260 260 260 260 260 260 260 260 260 260
OWNER ----H05 H05 H05 H05 H05 H05 H05 H05 H05 H05
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CROSS JOIN
You can join a table to itself.
The result set contains 20 * 20 = 400 rows
SELECT * FROM Owner, Car WHERE HP > 250; CARID ----F01 F01 F01 … F02 F02 F02 … F03 F03 F03 … F20 ©
PLATENUMBER ----------HD-V 106 HD-V 106 HD-V 106 … HD-VW 4711 HD-VW 4711 HD-VW 4711 … HD-JA 1972 HD-JA 1972 HD-JA 1972 … ?
2013 SAP AG. All rights reserved.
BRAND ----Fiat Fiat Fiat … VW VW VW … BMW BMW BMW … Audi
COLOR ----red red red … black black black … blue blue blue … green
SELECT * FROM Owner CROSS JOIN Car WHERE HP > 250; HP --75 75 75 … 120 120 120 … 184 184 184 … 184
OWNER ----H06 H06 H06 … H03 H03 H03 … H03 H03 H03 … ?
CARID ----F01 F02 F03 … F01 F02 F03 … F01 F02 F03 … F20
PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 … HD-V 106 HD-VW 4711 HD-JA 1972 … HD-V 106 HD-VW 4711 HD-JA 1972 … ?
BRAND ----Fiat VW BMW … Fiat VW BMW … Fiat VW BMW … Audi
COLOR ----red black blue … red black blue … red black blue … green
HP --75 120 184 … 75 120 184 … 75 120 184 … 184
OWNER ----H06 H03 H03 … H06 H03 H03 … H06 H03 H03 … ? 217
CROSS JOIN
You can combine more than two tables. The result set contains 10 * 20 * 3 = 600 rows
SELECT * FROM Owner CROSS JOIN Car CROSS JOIN Stolen;
SELECT * FROM Owner, Car, Stolen; OWNERID ------H01 H01 H01 … H01 H01 H01 … H02 H02 H02 … H10
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NAME ---Ms T Ms T Ms T … Ms T Ms T Ms T … Ms U Ms U Ms U … Mr Z
BIRTHDAY ---------1934-06-20 1934-06-20 1934-06-20 … 1934-06-20 1934-06-20 1934-06-20 … 1966-05-11 1966-05-11 1966-05-11 … 1986-02-03
2013 SAP AG. All rights reserved.
CITY ---------Wiesloch Wiesloch Wiesloch … Wiesloch Wiesloch Wiesloch … Hockenheim Hockenheim Hockenheim … Ladenburg
CARID ----F01 F01 F01 … F02 F02 F02 … F01 F01 F01 … F20
PLATENUMBER ----------HD-V 106 HD-V 106 HD-V 106 … HD-VW 4711 HD-VW 4711 HD-VW 4711 … HD-V 106 HD-V 106 HD-V 106 … ?
BRAND ----Fiat Fiat Fiat … VW VW VW … Fiat Fiat Fiat … Audi
COLOR ----red red red … black black black … red red red … green
HP --75 75 75 … 120 120 120 … 75 75 75 … 184
OWNER ----H06 H06 H06 … H03 H03 H03 … H06 H06 H06 … ?
PLATENUMBER ----------HD-VW 1999 HD-V 106 HD-Y 333 … HD-VW 1999 HD-V 106 HD-Y 333 … HD-VW 1999 HD-V 106 HD-Y 333 … HD-Y 333
REGISTERED_AT ------------2012-06-20 2012-06-01 2012-05-21 … 2012-06-20 2012-06-01 2012-05-21 … 2012-06-20 2012-06-01 2012-05-21 … 2012-05-21
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Joining Tables
INNER JOIN
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INNER JOIN
One row of the left table and one row of the right table are always joined to a common result row - provided that the JOIN condition is fulfilled. JOIN Condition: L.X
L
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… … … … … …
X 1 2 3 4 5
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… … … … … …
= R.Y
… … … … … …
R
… … … … … …
Y 3 4 5 6 7
… … … … … …
… … … … … …
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INNER JOIN
SELECT Column, Column, Column FROM Table, Table WHERE JOIN_Condition AND Supplementary_Condition; SELECT Column, Column, Column FROM Table JOIN Table ON JOIN_Condition WHERE Supplementary_Condition;
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Implicit INNER JOIN One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled. The JOIN condition is part of the WHERE clause To whom which car is registered?
SELECT * FROM Owner, Car WHERE OwnerID = Owner; OWNERID ------H06 H03 H03 H07 H03 …
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NAME -----Ms W SAP AG SAP AG IKEA SAP AG …
BIRTHDAY ---------1957-06-01 ? ? ? ? …
2013 SAP AG. All rights reserved.
CITY -------Wiesloch Walldorf Walldorf Walldorf Walldorf …
CARID ----F01 F02 F03 F04 F05 …
PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 …
BRAND -------Fiat VW BMW Mercedes Mercedes …
COLOR ----red black blue white black …
HP --75 120 184 136 170 …
OWNER ----H06 H03 H03 H07 H03 …
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Explicit INNER JOIN
One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled. The JOIN condition is part of the JOIN operation (and not of the WHERE clause) To whom is which car registered?
SELECT * FROM Owner INNER JOIN Car ON OwnerID = Owner; OWNERID ------H06 H03 H03 H07 H03 …
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NAME -----Ms W SAP AG SAP AG IKEA SAP AG …
BIRTHDAY ---------1957-06-01 ? ? ? ? …
2013 SAP AG. All rights reserved.
CITY -------Wiesloch Walldorf Walldorf Walldorf Walldorf …
CARID ----F01 F02 F03 F04 F05 …
PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 …
BRAND -------Fiat VW BMW Mercedes Mercedes …
COLOR ----red black blue white black …
HP --75 120 184 136 170 …
OWNER ----H06 H03 H03 H07 H03 …
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Explicit INNER JOIN
You can simply write "JOIN“ instead of "INNER JOIN“. The INNER JOIN is the most important JOIN variant (and therefore the default)
To whom is which car registered?
SELECT * FROM Owner JOIN Car ON OwnerID = Owner; OWNERID ------H06 H03 H03 H07 H03 …
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NAME -----Ms W SAP AG SAP AG IKEA SAP AG …
BIRTHDAY ---------1957-06-01 ? ? ? ? …
2013 SAP AG. All rights reserved.
CITY -------Wiesloch Walldorf Walldorf Walldorf Walldorf …
CARID ----F01 F02 F03 F04 F05 …
PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 …
BRAND -------Fiat VW BMW Mercedes Mercedes …
COLOR ----red black blue white black …
HP --75 120 184 136 170 …
OWNER ----H06 H03 H03 H07 H03 …
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Implicit INNER JOIN
You can specify certain columns in the projection list also for the implicit JOIN. To whom is which car registered?
SELECT Name, Brand, Color FROM Owner, Car WHERE OwnerID = Owner;
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NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Audi Renault VW
COLOR -----red black blue white black yellow blue black red black green red red white black orange red black
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Explicit INNER JOIN
You can specify certain columns in the projection list also for the explicit JOIN. To whom is which car registered?
SELECT Name, Brand, Color FROM Owner JOIN Car ON OwnerID = Owner;
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NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Audi Renault VW
COLOR -----red black blue white black yellow blue black red black green red red white black orange red black
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Implicit INNER JOIN
You can also use table aliases or tuple variables in the projection list for the implicit JOIN.
The JOIN condition can also refer to table aliases or tuple variables. If column names are not unique, you must qualify them NAME with the table aliases or tuple variables. -----Ms W To whom is which car registered? SAP AG
SELECT o.Name, c.Brand, c.Color FROM Owner o, Car c WHERE o.OwnerID = c.Owner;
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2013 SAP AG. All rights reserved.
SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Audi Renault VW
COLOR -----red black blue white black yellow blue black red black green red red white black orange red black
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Explicit INNER JOIN
You can use table aliases or tuple variables in the projection list for the explicit JOIN.
The JOIN condition can also refer to table aliases or tuple variables. If column names are not unique, you must qualify them NAME with the table aliases or tuple variables. -----Ms W To whom is which car registered? SAP AG
SELECT o.Name, c.Brand, c.Color FROM Owner o JOIN Car c ON o.OwnerID = c.Owner;
©
2013 SAP AG. All rights reserved.
SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Audi Renault VW
COLOR -----red black blue white black yellow blue black red black green red red white black orange red black
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Implicit INNER JOIN
You do not have to include a column from every involved table in the projection list.
Besides the JOIN condition the WHERE clause can contain additional conditions. To whom is a black car registered (at least one)?
SELECT DISTINCT o.Name FROM Owner o, Car c WHERE o.OwnerID = c.Owner AND c.Color = 'black';
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NAME -----Ms T SAP AG HDM AG IKEA
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Explicit INNER JOIN
You do not need to include a column from every involved table in the
projection list. In addition to the JOIN condition you can specify a WHERE clause. To whom is a black car registered (at least one)?
SELECT DISTINCT o.Name FROM Owner o JOIN Car c ON o.OwnerID = c.Owner WHERE c.Color = 'black';
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NAME -----Ms T SAP AG HDM AG IKEA
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Implicit INNER JOIN
You can explicitly rename result columns for the implicit JOIN. Sorting the result table is also possible To whom is a black car registered (at least one)?
SELECT DISTINCT o.Name AS "Owners' name" FROM Owner o, Car c WHERE o.OwnerID = c.Owner AND c.Color = 'black' ORDER BY o.Name; Owners'
name -----------HDM AG IKEA Ms T SAP AG
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Explicit INNER JOIN
You can explicitly rename result columns for the explicit JOIN. Sorting the result table is also possible To whom is a black car registered (at least one)?
SELECT DISTINCT o.Name AS "Owners' name" FROM Owner o JOIN Car c ON o.OwnerID = c.Owner WHERE c.Color = 'black' ORDER BY o.Name; Owners'
name -----------HDM AG IKEA Ms T SAP AG
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Implicit INNER JOIN
You can add a GROUP BY clause for the implicit Join. How many black cars are registered to which owner?
SELECT FROM WHERE GROUP BY ORDER BY
o.Name, COUNT(*) AS "# black cars" Owner o, Car c o.OwnerID = c.Owner AND c.Color = 'black' o.OwnerID, o.Name 2 DESC, 1 ASC; NAME -----SAP AG HDM AG IKEA Ms T
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# black cars -----------3 1 1 1
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Explicit INNER JOIN
You can add a GROUP BY clause for the explicit Join. How many black cars are registered to which owner?
SELECT o.Name, COUNT(*) AS "# black cars" FROM Owner o JOIN Car c ON o.OwnerID = c.Owner WHERE c.Color = 'black' GROUP BY o.OwnerID, o.Name ORDER BY 2 DESC, 1 ASC; NAME # black -----SAP AG HDM AG IKEA Ms T
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cars -----------3 1 1 1
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Implicit INNER JOIN
You can build the JOIN condition on multiple columns for the implicit JOIN. To whom within the EU is a black car registered (at least one)?
SELECT DISTINCT o.Name AS "Owners' name" FROM Owner_EU o, Car_EU c WHERE o.Country = c.Country AND o.OwnerID = c.Owner AND c.Color = 'black' Owners' name ORDER BY o.Name; -----------HDM AG IKEA Ms O Ms T SAP AG Señora R
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Explicit INNER JOIN
You can build the JOIN condition on multiple columns for the explicit JOIN. To whom within the EU is a black car registered (at least one)?
SELECT DISTINCT o.Name AS "Owners' name" FROM Owner_EU o JOIN Car_EU c ON o.Country = c.Country AND o.OwnerID = c.Owner WHERE c.Color = 'black' Owners' name ORDER BY o.Name; -----------HDM AG IKEA Ms O Ms T SAP AG Señora R
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Implicit INNER JOIN
You can join a table to itself. Who is the manager of which employee?
SELECT e.Name AS Employee, m.Name AS Manager FROM Official e, Official m WHERE e.Manager = m.PNr; EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H
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MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I
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Explicit INNER JOIN
You can join a table to itself also for the explicit JOIN. Who is the manager of which employee?
SELECT e.Name AS Employee, m.Name AS Manager FROM Official e JOIN Official m ON e.Manager = m.PNr; EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H
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MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I
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Implicit INNER JOIN
You can combine more than two tables. To whom is which stolen car registered?
SELECT o.Name, c.Brand, c.Color, c.PlateNumber FROM Owner o, Car c, Stolen s WHERE o.OwnerID = c.Owner AND c.PlateNumber = s.PlateNumber; NAME ---Ms W Mr V Ms Y
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BRAND ----Fiat Audi Audi
COLOR -----red yellow orange
PLATENUMBER ----------HD-V 106 HD-VW 1999 HD-Y 333
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Explicit INNER JOIN
You can combine more than two tables also for the explicit JOIN. To whom is which stolen car registered?
SELECT o.Name, c.Brand, c.Color, c.PlateNumber FROM Owner o JOIN Car c ON o.OwnerID = c.Owner JOIN Stolen s ON c.PlateNumber = s.PlateNumber;
NAME ---Ms W Mr V Ms Y
©
2013 SAP AG. All rights reserved.
BRAND ----Fiat Audi Audi
COLOR -----red yellow orange
PLATENUMBER ----------HD-V 106 HD-VW 1999 HD-Y 333
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Explicit INNER JOIN
You can use different comparison operators other than EQUAL in the JOIN condition. Which owner is older than which other owner?
SELECT o.Name AS "older", y.Name AS "younger" FROM Owner o JOIN Owner y ON o.Birthday < y.Birthday;
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older ----Ms T Ms T Ms T Ms T Ms T Ms T Ms U Ms U Ms U Mr V Mr V Mr V Mr V Mr V Ms W Ms W Ms W Ms W Ms Y Mr Z Mr Z
younger ------Ms U Mr V Ms W Mr X Ms Y Mr Z Mr X Ms Y Mr Z Ms U Ms W Mr X Ms Y Mr Z Ms U Mr X Ms Y Mr Z Mr X Mr X Ms Y
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Explicit INNER JOIN
You can use calculations in the JOIN condition. Which car has at least three times the power than which other car?
SELECT m.*, l.* FROM Car m JOIN Car l ON m.HP >= l.HP * 3; CARID ----F06
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PLATENUMBER ----------HD-VW 1999
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BRAND ----Audi
COLOR -----yellow
HP --260
OWNER ----H05
CARID ----F01
PLATENUMBER ----------HD-V 106
BRAND ----Fiat
COLOR ----red
HP -75
OWNER ----H06
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Explicit INNER JOIN
You can use functions in the JOIN condition. Which owner was born in the same year as which other owner?
SELECT o1.Name, o1.Birthday, o2.Name, o2.Birthday FROM Owner o1 JOIN Owner o2 ON YEAR(o1.Birthday) = YEAR(o2.Birthday) AND o1.OwnerID < o2.OwnerID;
NAME ---Mr X Mr X Ms Y
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BIRTHDAY ---------1986-08-30 1986-08-30 1986-02-10
NAME ---Ms Y Mr Z Mr Z
BIRTHDAY ---------1986-02-10 1986-02-03 1986-02-03
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Joining Tables
OUTER JOIN
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OUTER JOIN
The OUTER JOIN has three sub types: LEFT OUTER JOIN RIGHT OUTER JOIN FULL OUTER JOIN
For all three sub types of the OUTER JOIN SAP HANA only provides the explicit syntax variant.
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LEFT OUTER JOIN
One row of a table and one row of another table are always connected to a common result row - provided the JOIN condition is fulfilled. In addition, rows of the left table without matching row in the right table are copied to the query result. The missing values (from the right table) are filled with NULL values.
L
©
… … … … … …
X 1 2 3 4 5
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… … … … … …
… … … … … …
NULL NULL NULL NULL
R
… … … … … …
Y 3 4 5 6 7
… … … … … …
… … … … … … 246
RIGHT OUTER JOIN
One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled. In addition, rows of the right table without matching row in the left table are copied to the query result. The missing values (from the left table) are filled with NULL values.
L
… … … … … …
X 1 2 3 4 5
… … … … … …
… … … … … …
NULL NULL NULL NULL
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2013 SAP AG. All rights reserved.
R
… … … … … …
Y 3 4 5 6 7
… … … … … …
… … … … … … 247
FULL OUTER JOIN
One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled. In addition, rows of both tables without matching records are copied to the query result. The missing values (from the other table) are filled with NULL values.
L
… … … … … …
X 1 2 3 4 5
… … … … … …
… … … … … …
NULL NULL NULL NULL
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2013 SAP AG. All rights reserved.
NULL NULL NULL NULL
R
… … … … … …
Y 3 4 5 6 7
… … … … … …
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OUTER JOIN
SELECT Column, Column, Column FROM Table LEFT OUTER JOIN Table ON JOIN_Condition WHERE Additional_Condition; SELECT Column, Column, Column FROM Table RIGHT OUTER JOIN Table ON JOIN_Condition WHERE Additional_Condition; SELECT Column, Column, Column FROM Table FULL OUTER JOIN Table ON JOIN_Condition WHERE Additional_Condition;
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LEFT OUTER JOIN
Which car is registered to which individual or company? Individuals and companies that currently do not have a car registered should be included. Cars without a registration should not be included in the result.
SELECT Name, CarID, Brand FROM Owner LEFT OUTER JOIN Car ON OwnerID = Owner;
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2013 SAP AG. All rights reserved.
NAME -----Ms T Ms U Ms U SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG HDM AG HDM AG Mr V Ms W IKEA IKEA IKEA Mr X Ms Y Mr Z
CARID ----F19 F09 F11 F02 F03 F18 F05 F15 F07 F14 F12 F10 F06 F01 F13 F08 F04 ? F17 ?
BRAND -------VW Skoda BMW VW BMW Renault Mercedes Skoda Audi Mercedes Skoda BMW Audi Fiat Renault VW Mercedes ? Audi ?
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RIGHT OUTER JOIN
Which car is registered to which individual or company? Cars without a registration should be included. Individuals and companies that currently do not have a car registered should not be included in the result.
SELECT Name, CarID, Brand FROM Owner RIGHT OUTER JOIN Car ON OwnerID = Owner;
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2013 SAP AG. All rights reserved.
NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ?
CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi
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FULL OUTER JOIN
Which car is registered to which individual or company? Individuals and companies that currently do not have a car registered should be included. Cars without a registration should be included in the result .
SELECT Name, CarID, Brand FROM Owner FULL OUTER JOIN Car ON OwnerID = Owner;
©
2013 SAP AG. All rights reserved.
NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ? Mr X Mr Z
CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 ? ?
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi ? ?
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LEFT OUTER JOIN
For the LEFT OUTER JOIN you can also add a WHERE clause in addition to the JOIN condition. Which car is registered to which individual? Assume that companies don’t have a birthday. Individuals that currently do not have a car registered should be included. Cars without a registration should not be included in the result.
SELECT Name, CarID, Brand FROM Owner LEFT OUTER JOIN Car ON OwnerID = Owner WHERE Birthday IS NOT NULL;
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2013 SAP AG. All rights reserved.
NAME -----Ms T Ms U Ms U SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG HDM AG HDM AG Mr V Ms W IKEA IKEA IKEA Mr X Ms Y Mr Z
CARID ----F19 F09 F11 F02 F03 F18 F05 F15 F07 F14 F12 F10 F06 F01 F13 F08 F04 ? F17 ?
BRAND -------VW Skoda BMW VW BMW Renault Mercedes Skoda Audi Mercedes Skoda BMW Audi Fiat Renault VW Mercedes ? Audi ?
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RIGHT OUTER JOIN
For the RIGHT OUTER JOIN you can also add a WHERE clause in addition to the JOIN condition. Which car is registered to which individual? Individuals who currently do not have a car registered should be excluded. (Realized by RIGHT OUTER JOIN) Cars without a registration should not be included in the result. (Realized by the WHERE clause)
SELECT Name, CarID, Brand FROM Owner RIGHT OUTER JOIN Car ON OwnerID = Owner WHERE Birthday IS NOT NULL;
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2013 SAP AG. All rights reserved.
NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ?
CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi
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FULL OUTER JOIN
For the FULL OUTER JOIN you can also add a WHERE clause in addition to the JOIN condition. Which car is registered to which individual? Individuals that currently do not have a car registered should be included. Cars without a registration should not be included in the result. (Realized by the WHERE clause)
SELECT Name, CarID, Brand FROM Owner FULL OUTER JOIN Car ON OwnerID = Owner WHERE Birthday IS NOT NULL;
©
2013 SAP AG. All rights reserved.
NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ? Mr X Mr Z
CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 ? ?
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi ? ?
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LEFT OUTER JOIN
You can LEFT OUTER JOIN a table with itself Who is the manager of which employee? Employees without a manager should be included. Only Managers with employees should be included.
EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I
MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I ?
SELECT e.Name AS Employee, m.Name AS Manager FROM Official e LEFT OUTER JOIN Official m ON e.Manager = m.PNr;
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RIGHT OUTER JOIN
You can RIGHT OUTER JOIN a table with itself Who is the manager of which employee? The MANAGER column should also include officials that are not managing employees. The EMPLOYEE column should only include officials with a manager assigned.
SELECT e.Name AS Employee, m.Name AS Manager FROM Official e RIGHT OUTER JOIN Official m ON e.Manager = m.PNr;
©
2013 SAP AG. All rights reserved.
EMPLOYEE -------? ? ? Mr A Mr B Ms C ? ? ? Mr E Mr F Ms G Ms D Ms H
MANAGER ------Mr A Mr B Ms C Ms D Ms D Ms D Mr E Mr F Ms G Ms H Ms H Ms H Mr I Mr I
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FULL OUTER JOIN
You can FULL OUTER JOIN a table with itself Who is the manager of which employee? The EMPLOYEE column should include officials without a manager. The MANAGER column should also include officials that are not managing employees.
SELECT e.Name AS Employee, m.Name AS Manager FROM Official e FULL OUTER JOIN Official m ON e.Manager = m.PNr;
©
2013 SAP AG. All rights reserved.
EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I ? ? ? ? ? ?
MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I ? Mr A Mr B Ms C Mr E Mr F Ms G
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LEFT/RIGHT/FULL OUTER JOIN You can also … Project to certain columns Reference table aliases or tuple variables in the projection list. Use table aliases or tuple variables in the JOIN condition. Explicitly rename result columns. Sort the query result. Add a GROUP BY clause. Use aggregate expressions. eliminate duplicates by using DISTINCT. Specify a JOIN condition on multiple columns. Use any comparison operator in the JOIN condition. JOIN a table with itself. JOIN more than two tables. ©
2013 SAP AG. All rights reserved.
and: You have to qualify column names with the table alias or with a tuple variable if the column names are not unique. You do not need to include a column from every table involved in the JOIN into the projection list.
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Unit 4: Reading Data From Multiple Tables Part I Summary
You should now be able to: •
Merge the result of several select statements using the UNION statement • Combine data from several tables when querying data using JOIN constructs • •
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List the various types of Joins Explain the differences between the various types of Joins, and apply the right type of Join depending on the problem
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260
Unit 5 Reading Data From Multiple Tables Part II
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Unit 5: Reading Data From Multiple Tables Part II Learning Objectives After completing this unit, you will be able to: • Use sub-queries to query data from multiple tables in a single select statement
• Explain the difference between uncorrelated and correlated subqueries
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262
Nested Queries
Nested Queries
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Nested Queries
In addition to UNION and JOIN, Nested Queries also provide the option of reading from multiple tables and views. Nested Queries contain a so called “Sub Query”.
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Sub Query
What is a sub query?
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Sub Query
A sub query is a query (SELECT-FROM-WHERE statement) that is used in another query (SELECT-FROM-WHERE statement). The "other Query" – containing the sub query - is called the "outer query". In this context the Sub Query is also referred to as "inner query".
Outer Query
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SELECT PlateNumber, Brand, Color Sub Query FROM Car WHERE Owner IN ( SELECT OwnerID FROM Owner WHERE City = 'Wiesloch');
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Sub Query
When is a sub query useful?
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Sub Query
A sub query is useful if ... It makes the SELECT statement more "readable" - for example, because the use of nested queries is the "obvious way" The SELECT statement has a better performance – however, with a "perfect" optimizer, this should not be the case. The formulation of the SELECT statement is not possible (or only extremely cumbersome) without a sub query - which can be particularly the case for aggregate expressions.
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Sub Query
What is the difference between an uncorrelated and a correlated sub query?
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269
Sub Query
An uncorrelated sub query makes no reference to the outer query.
SELECT * FROM Car WHERE Owner IN (SELECT OwnerID FROM Owner WHERE City = 'Wiesloch'); A correlated sub query refers to the outer query. SELECT * FROM Car c WHERE EXISTS (SELECT * FROM Owner o WHERE o.OwnerID = c.Owner AND o.City = 'Wiesloch');
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Sub Query
Uncorrelated sub queries
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Uncorrelated Sub Query
SELECT Column, Column, Column FROM Table WHERE Column IN (SELECT Column FROM Table WHERE Condition);
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Uncorrelated Sub Query
You can use IN, if the outer value should be included in the result set of the sub query. Which cars are registered to an owner from Wiesloch?
SELECT * FROM Car WHERE Owner IN (SELECT OwnerID FROM Owner WHERE City = 'Wiesloch');
CARID ----F01 F19
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2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-V 106 HD-VW 2012
BRAND ----Fiat VW
COLOR ----red black
HP --75 125
OWNER ----H06 H01
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Uncorrelated Sub Query
You can use =
ANY, if the outer value should match any result value of the sub query.
Which cars are registered to an owner from Wiesloch?
SELECT * FROM Car WHERE Owner = ANY (SELECT OwnerID FROM Owner WHERE City = 'Wiesloch');
CARID ----F01 F19
©
2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-V 106 HD-VW 2012
BRAND ----Fiat VW
COLOR ----red black
HP --75 125
OWNER ----H06 H01
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Uncorrelated Sub Query
“= ANY” is equivalent to “IN”. So, where is the value of “= ANY” ?
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Uncorrelated Sub Query
= ANY is equivalent to IN, but you can use ANY with other comparison operators: = ANY
The external value matches any value of the sub query.
< ANY
The external value is less than any value of the sub query.
ANY
©
The external value is less or equal than any value of the sub query. The external value is greater than any value of the sub query.
>= ANY
The external value is greater or equal than any value of the sub query.
ANY
The external value is different to any value of the sub query.
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276
Uncorrelated Sub Query
You can use > ANY, if the external value should be greater than any value of the sub query.
Which officials have more than the minimum overtime hours?
SELECT Name, Overtime FROM Official WHERE Overtime > ANY (SELECT Overtime FROM Official); NAME ---Ms C Mr F Ms G
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OVERTIME -------20 18 22
277
Uncorrelated Sub Query
You can use > instead of > ANY, if the sub query results in only a single value.
Which officials have more than the minimum overtime hours?
SELECT Name, Overtime FROM Official WHERE Overtime > (SELECT MIN(Overtime) FROM Official); NAME ---Ms C Mr F Ms G
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2013 SAP AG. All rights reserved.
OVERTIME -------20 18 22
278
Uncorrelated Sub Query
You can use < ALL, if the outer value should be less than all result values of the sub query (other comparison operators are possible) = ALL
The external value matches each result value of the sub query. (This option is quite useful, as the sub query result can contain duplicates.)
< ALL
The external value is less than all values of the sub query.
ALL
©
The external value is less or equal than all values of the sub query. The external value is greater than all values of the sub query.
>= ALL
The external value is greater or equal than all values of the sub query.
ALL
The external value is different to all values of the sub query.
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279
Uncorrelated Sub Query
You can use 100 Power DESC, CarID ASC;
CARID ----F13 F09 F12
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PLATENUMBER ----------HD-IK 1001 HD-UP 13 HD-XY 4711
POWER ----136 105 105
416
VIEWs
You can INSERT new rows into a view.
The new rows are not inserted into the view, but into the underlying base table!
INSERT INTO redCars VALUES ('F77', 'red', 'HD-MT 2509', 170);
SELECT * FROM Car; CARID ----F01 F02 F03 … F18 F19 F20 F77
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2013 SAP AG. All rights reserved.
PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 … HD-MQ 2006 HD-VW 2012 ? HD-MT 2509
BRAND ------Fiat VW BMW … Renault VW Audi ?
COLOR ----red black blue … red black green red
HP --75 120 184 … 90 125 184 170
OWNER ----H06 H03 H03 … H03 H01 ? ? 417
VIEWs You can UPDATE rows in a view.
The rows are not changed in the view, but in the underlying base table!
UPDATE redCars SET Power = 120 WHERE Power < 120; SELECT * FROM Car;
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CARID ----F01 … F07 F08 F09 F10 F11 F12 F13 … F18 …
PLATENUMBER ----------HD-V 106 … HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 … HD-MQ 2006 …
BRAND ------Fiat … Audi VW Skoda BMW BMW Skoda Renault … Renault …
COLOR ----red … blue black red black green red red … red …
HP --120 … 116 160 120 140 184 120 136 … 120 …
OWNER ----H06 … H03 H07 H02 H04 H02 H04 H07 … H03 … 418
VIEWs You can DELETE rows from a view.
The rows are not deleted from the view, but from the underlying base table!
DELETE FROM redCars WHERE Power < 125; SELECT * FROM Car;
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2013 SAP AG. All rights reserved.
CARID ----F01 … F07 F08 F09 F10 F11 F12 F13 … F18 …
PLATENUMBER ----------HD-V 106 … HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 … HD-MQ 2006 …
BRAND ------Fiat … Audi VW Skoda BMW BMW Skoda Renault … Renault …
COLOR ----red … blue black red black green red red … red …
HP --120 … 116 160 120 140 184 120 136 … 120 …
OWNER ----H06 … H03 H07 H02 H04 H02 H04 H07 … H03 … 419
Data Definition
Which SQL statements can I use to create or delete views?
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420
Data Definition
CREATE VIEW Creates a new view.
DROP VIEW Deletes an existing view.
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421
Data Definition
CREATE VIEW
©
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422
CREATE VIEW
CREATE VIEW SELECT FROM WHERE
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View AS Column, Column, Column Table Condition;
423
CREATE VIEW
You can restrict a view to specific columns of the underlying base table.
CREATE VIEW Autos AS SELECT Brand, Color, HP FROM Car;
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2013 SAP AG. All rights reserved.
BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi
COLOR -----red black blue white black yellow blue black red black green red red white black green orange red black green
HP --75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184
424
CREATE VIEW
You can restrict a view to specific rows of the underlying base table.
CREATE VIEW SELECT FROM WHERE
Audis AS Brand, Color, HP Car Brand = 'Audi';
BRAND ----Audi Audi Audi Audi
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2013 SAP AG. All rights reserved.
COLOR -----yellow blue orange green
HP --260 116 184 184
425
CREATE VIEW
You can restrict a view to a certain amount of rows:
CREATE VIEW SELECT FROM WHERE
"Top3 Audis" AS TOP 3 Brand, Color, HP Car Brand = 'Audi';
BRAND ----Audi Audi Audi
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2013 SAP AG. All rights reserved.
COLOR -----yellow blue orange
HP --260 116 184
426
CREATE VIEW
The columns of a view can have different names than the columns of the underlying base table.
CREATE VIEW SELECT FROM WHERE
Audis (Manufacturer, Color, Power) AS Brand, Color, HP Car Brand = 'Audi';
MANUFACTURER -----------Audi Audi Audi Audi
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2013 SAP AG. All rights reserved.
COLOR -----yellow blue orange green
POWER ----260 116 184 184
427
CREATE VIEW
You can (and should) rename the columns in the SELECT clause
CREATE VIEW SELECT FROM WHERE
Audis AS Brand AS Manufacturer, Color, HP AS Power Car Brand = 'Audi';
MANUFACTURER -----------Audi Audi Audi Audi
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2013 SAP AG. All rights reserved.
COLOR -----yellow blue orange green
POWER ----260 116 184 184
428
CREATE VIEW
If you rename columns in both the CREATE VIEW clause and in the SELECT clause, the name of the CREATE VIEW clause "wins“. We recommend renaming columns only in the SELECT clause.
CREATE VIEW SELECT FROM WHERE
Audis (Manufacturer, Color, Power) AS Brand AS Description, Color, HP AS HorsePower Car Brand = 'Audi';
MANUFACTURER -----------Audi Audi Audi Audi
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COLOR -----yellow blue orange green
POWER ----260 116 184 184
429
CREATE VIEW
A view can contain duplicates.
CREATE VIEW Colors AS SELECT Color FROM Car;
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COLOR -----red black blue white black yellow blue black red black green red red white black green orange red black green
430
CREATE VIEW
You can use DISTINCT in the view definition. Thus the view does not contain duplicates.
CREATE VIEW Colors AS SELECT DISTINCT Color FROM Car;
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COLOR -----red black blue white yellow green orange
431
CREATE VIEW
You can use ORDER BY in the view definition. From a theoretical perspective the use of ORDER BY is highly questionable, because the rows of a view are unsorted! There is no guarantee that the specified ORDER BY sorting is taken into account when reading the view. We recommend avoiding ORDER BY in the view definition.
CREATE VIEW SELECT FROM ORDER BY
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Colors AS DISTINCT Color Car Color;
COLOR -----blue yellow green orange red black white 432
CREATE VIEW
You can use calculated columns in a view.
CREATE VIEW SELECT FROM WHERE
Audis AS CarID, Color, ROUND(HP / 1.35962162, 0) AS "kW" Car Brand = 'Audi';
CARID ----F06 F07 F17 F20
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COLOR -----yellow blue orange green
kW --191 85 135 135
433
CREATE VIEW
You can use functions in the view definition.
CREATE VIEW OwnerOverview AS SELECT Name, YEAR(Birthday) AS YearOfBirth FROM Owner; NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z ©
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YEAROFBIRTH ----------1934 1966 ? ? 1952 1957 ? 1986 1986 1986 434
CREATE VIEW
You can use GROUP BY and aggregate expressions in the view definition.
CREATE VIEW SELECT FROM GROUP BY HAVING
Cars AS Brand, COUNT(*) AS Quantity Car Brand COUNT(*) > 2; BRAND -------VW BMW Mercedes Audi Skoda
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QUANTITY -------3 3 3 4 3
435
CREATE VIEW
You can use UNION [ALL] in the view definition.
CREATE VIEW Persons AS SELECT OwnerID AS "Pers ID", Name FROM Owner WHERE Birthday >= '1977-05-21' UNION ALL SELECT PNr, Name FROM Official WHERE Salary = 'A09';
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Pers ID ------H08 H09 H10 P01 P03 P06
NAME ---Mr X Ms Y Mr Z Mr A Ms C Mr F
436
CREATE VIEW
You can use (any) JOIN in the view definition.
CREATE VIEW SELECT FROM WHERE
StolenCars AS c.PlateNumber, c.Brand, c.Color Car c, Stolen s c.PlateNumber = s.PlateNumber;
PLATENUMBER ----------HD-V 106 HD-VW 1999 HD-Y 333
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2013 SAP AG. All rights reserved.
BRAND ----Fiat Audi Audi
COLOR -----red yellow orange
437
CREATE VIEW
You can use an uncorrelated sub query in the view definition.
CREATE VIEW SELECT FROM WHERE
StolenCars AS PlateNumber, Brand, Color Car PlateNumber IN (SELECT PlateNumber FROM Stolen);
PLATENUMBER ----------HD-V 106 HD-VW 1999 HD-Y 333
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2013 SAP AG. All rights reserved.
BRAND ----Fiat Audi Audi
COLOR -----red yellow orange
438
CREATE VIEW
You can use a correlated sub query in the view definition.
CREATE VIEW SELECT FROM WHERE
StolenCars AS c.PlateNumber, Car c EXISTS (SELECT FROM WHERE
c.Brand, c.Color * Stolen s s.PlateNumber = c.PlateNumber);
PLATENUMBER ----------HD-V 106 HD-VW 1999 HD-Y 333
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2013 SAP AG. All rights reserved.
BRAND ----Fiat Audi Audi
COLOR -----red yellow orange
439
CREATE VIEW
You can define a view based on another view.
CREATE VIEW SELECT FROM WHERE
Volkswagen AS CarID, PlateNumber, Color, HP Car Brand = 'VW';
CREATE VIEW SELECT FROM WHERE
blackVolkswagen AS CarID, PlateNumber, HP Volkswagen Color = 'black';
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2013 SAP AG. All rights reserved.
CARID ----F02 F08 F19
PLATENUMBER ----------HD-VW 4711 HD-IK 1002 HD-VW 2012
HP --120 160 125
440
Data Definition
DROP VIEW
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441
DROP VIEW
DROP VIEW View;
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442
DROP VIEW
You can drop an existing view. Only the view definition is deleted. The underlying data of the view is not deleted.
DROP VIEW redCars;
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443
Views and Changeability
Is it always possible to change the data of the underlying table by referencing a view?
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444
Views and Changeability
Referencing a view, the rows of the underlying base table cannot be changed if (at minimum) any of the following statements is true: The view definition contains a calculation in the projection list (such as HP / 1.36). The view definition uses a function in the projection list (such as YEAR). The view definition uses DISTINCT to eliminate duplicates. The view definition contains the TOP n clause to restrict the number of rows to n. The view definition uses an aggregate function in the projection list (such as MAX). The view definition is based on a GROUP BY clause.
The view definition uses UNION. The view definition contains an (implicit or explicit) JOIN. The view definition uses a sub query in the projection list.
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445
Unit 9: Using Views For Data Access Summary
You should now be able to: • Describe the use cases for and advantages of using database views • Define database views using SQL • Query data through views
• Delete database views using SQL
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446
Unit 10 Defining Data Access
447
Unit 10: Defining Data Access Learning Objectives
After completing this unit, you will be able to: • Access tables in other schemas (assuming suitable permissions) • Control how other users can access data in your own schema
• Explain when database indexes make sense even in HANA • Create and delete database indexes using SQL
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448
Data Definition
Why are there no naming collisions when multiple users create the same table or view?
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449
Schemata
Please note: The following information applies to direct (SQL-based) work on SAP HANA - but not to database access using ABAP! User 1 creates table “Car”. User 2 also creates table “Car”. Why is there no naming collision? The name of the database object (table, view etc.) implicitly contains a schema name as prefix: . “USER1.Car” ≠ “USER2.Car”
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450
Schemata
When creating a user an identically named schema is created implicitly. By default a user works in his own schema (and namespace). You can explicitly specify the schema name if necessary. Which brand has (at least) one black car? SELECT FROM WHERE ORDER BY
© 2013
DISTINCT User1.Car.Brand User1.Car User1.Car.Color = 'black' User1.Car.Brand;
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BRAND -------BMW Mercedes Skoda VW 451
Schemata
Queries across multiple schemata are possible. To whom is (at least) one black car registered? SELECT DISTINCT Schema1.Owner.Name FROM Schema1.Owner, Schema2.Car WHERE Schema1.Owner.OwnerID = Schema2.Car.Owner AND Schema2.Car.Color = 'black';
NAME -----SAP AG IKEA HDM AG Ms T © 2013
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452
Access Control
How can I define who has access to which data?
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453
Access Control
To specify who can access which data, you can use the following two main (SQL based) options. They can be used in combination: Create views that represent a portion of the data: – You provide a view that contains only the owners of CityA to a user for whom the owners outside of CityA are not relevant.
Grant specific access permissions to selected users: – You grant read only permissions to a user, who should be able to read the owner data but not to change it.
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454
Access Control
Which SQL statements can I use for access control?
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455
Access Control
GRANT You can selectively grant access permissions to a user (or role).
REVOKE You can selectively revoke access permissions from a user (or role).
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456
Access Control
The SQL statement GRANT
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457
GRANT
GRANT Privilege, Privilege ON Database_Object TO Database_User WITH GRANT OPTION;
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458
GRANT
You can define which user is allowed to SELECT from which table.
By adding the WITH GRANT OPTION the recipient of the privilege (grantee) is allowed to grant this privilege to other users (without losing it).
User1 is allowed to read (SELECT) from table Official of schema Schema2. User1 is allowed to grant this SELECT privilege to other users.
GRANT ON TO WITH
© 2013
SELECT Schema2.Official User1 GRANT OPTION;
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459
GRANT
You can exactly specify which user is allowed to INSERT rows into which table.
The INSERT privilege does not necessarily require the SELECT privilege.
The INSERT privilege does not automatically include the SELECT privilege.
User1 is allowed to INSERT rows into table Owner of schema Schema2. As User 1 has no SELECT privilege, he is not allowed to read the rows he has inserted.
GRANT INSERT ON Schema2.Owner TO User1;
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460
GRANT
You can exactly specify which user is allowed to UPDATE rows of which table.
The UPDATE privilege does not necessarily require the SELECT privilege.
The UPDATE privilege does not automatically include the SELECT privilege.
User1 is allowed to UPDATE table Car_EU of schema Schema2. Because User1 has no SELECT privileges for the table Car_EU the WHERE clause of his UPDATE statement cannot contain a reference to the columns of this table.
GRANT UPDATE ON Schema2.Car_EU TO User1;
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461
GRANT
You can exactly specify which user is allowed to DELETE rows from which table.
The DELETE privilege does not necessarily require the SELECT privilege.
The DELETE privilege does not automatically include the SELECT privilege.
User1 is allowed to DELETE rows from table Stolen of schema Schema2. Because User1 has no SELECT privileges for the table Stolen the WHERE clause of his DELETE statement cannot contain a reference to the columns of this table.
GRANT DELETE ON Schema2.Stolen TO User1;
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462
GRANT
You can assign multiple privileges for the same database object by using a single GRANT statement. User1 is allowed to SELECT, INSERT, UPDATE, DELETE on table Contact of schema Schema2. User1 is allowed to grant these privileges to other users (in whole or in part). GRANT ON TO WITH
© 2013
SELECT, INSERT, UPDATE, DELETE Schema2.Contact User1 GRANT OPTION;
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463
GRANT
You can assign privileges for all tables and views of a schema by using a single GRANT statement. User1 is allowed to SELECT from all tables and views of schema Schema2. User1 is allowed to grant this privilege to other users (for all or selected tables or views).
GRANT ON TO WITH
© 2013
SELECT SCHEMA Schema2 User1 GRANT OPTION;
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464
GRANT
Depending on the granularity (or the type of database object) you can assign different access privileges to users.
Granularity
Access Privileges
(single) Schema
SELECT, INSERT, UPDATE, DELETE, DROP, ALTER, CREATE ANY
(single) Table
SELECT, INSERT, UPDATE, DELETE, DROP, ALTER
(single) View
SELECT, INSERT, UPDATE, DELETE, DROP
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465
Access Control
The SQL statement REVOKE
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466
REVOKE
REVOKE Privilege, Privilege, Privilege ON Database_Object FROM Database_User;
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467
REVOKE
Using the REVOKE statement you can selectively revoke access privileges from a user.
User1 will no longer be allowed to read from table Official of schema Schema2 (assuming that this privilege is not granted by other privileges)
REVOKE SELECT ON Schema2.Official FROM User1;
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468
REVOKE
You can revoke multiple privileges for the same database object by using a single REVOKE statement
User1 will no longer be allowed to SELECT, INSERT, UPDATE, DELETE from table Contact of schema Schema2 (assuming that these privileges are not granted by other privileges)
REVOKE SELECT, INSERT, UPDATE, DELETE ON Schema2.Contact FROM User1;
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469
REVOKE
Access privileges on different hierarchical levels of database objects are independent.
Despite the REVOKE statement User1 still has read access to the table Official of schema Schema2 , because he still has a (not withdrawn) read permission on the entire schema Schema2 .
GRANT SELECT ON SCHEMA Schema2 TO User1; REVOKE SELECT ON Schema2.Official FROM User1;
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470
REVOKE
The independence of access privileges at various hierarchical levels is also valid in the contrary case.
Despite the REVOKE statement User1 still has read access to the table Official of schema Schema2 , because he still has a (not withdrawn) read permission on the table.
GRANT SELECT ON Schema2.Official TO User1; REVOKE SELECT ON SCHEMA Schema2 FROM User1;
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471
Access Paths
What is a database index?
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472
Access Paths
An index (plural: indexes or indices) is an access path. An index may speed up search and sorting. – Index access instead of "full table scan" or "full column scan" – Index usage with HANA COLUMN TABLES only in exceptional cases
An index usually slows down inserts, update and deletion. An index has no influence on the query result. – Effect "only" on performance
Indexes are not covered by the SQL standard. – CREATE / DROP INDEX are strictly speaking no SQL statements.
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473
HANA: Why Indexes on Column Store?
– Sorted dictionary of values – (Bit-)Vector of value IDs
Are indexes necessary to speed up data access?
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Column „Name“ conceptual
Column “Name” – internal representation Value ID vector
Dictionary
Miller
4
0
Baker
Jones
1
1
Jones
Millman
5
2
John
Zsuwalski
772
3
Johnson
Baker
0
4
Miller
Miller
4
5
Millman
John
2
Miller
4
Johnson
3
Jones
1
…
…
sorted
HANA stores data in columns, not rows Columns are stored as
… 772
Zsuwalski …
474
HANA: Table Scan With Value ID
Value ID lookup is fast
Dictionary
–
WHERE Name = 'Zsuwalski'
Row ID lookup can still involve “full column scan” – No issue for most value ID vectors, even less if sorted – But may be slow for very large value ID vectors
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binary search on dictionary
– binary search on sorted dictionary
Value ID vector
ID
Value
Row ID
Value ID
0
Baker
0
4
1
Jones
1
1
2
John
2
5
3
Johnson
3
772
4
Miller
4
0
5
Millman
772
full column scan with value ID
…
…
254621
772
Zsuwalski
…
…
…
475
HANA: Inverted Index on Column Store
An inverted index can be reasonable in exceptional cases if column scan performance is not sufficient
WHERE Name = 'Zsuwalski'
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binary search on dictionary
Dictionary
Inverted Index
Value ID vector
ID
Value
Value ID
Rows
Row ID
Value ID
0
Baker
0
56756
0
4
1
Jones
1
345, 76876
…
…
3
772
… 772
Zsuwalski …
…
…
772
3, 254621
…
…
… 254621
772
…
…
476
Access Paths
Which SQL statements can I use to create a database index?
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477
Access Paths
CREATE INDEX A new database index is created.
DROP INDEX An existing database index is deleted.
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478
Access Paths
CREATE INDEX
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479
Access Paths
CREATE INDEX Access_Path ON Table (Column, Column);
CREATE UNIQUE INDEX Access_Path ON Table (Column, Column);
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480
CREATE INDEX
To speed up the read access you can create an index on a table column. The respective table can be empty or contain data.
An index on column PlateNumber of table Car will be created:
CREATE INDEX PlateNumberIndex ON Car (PlateNumber);
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481
CREATE INDEX
You can create a UNIQUE INDEX to ensure that the corresponding column cannot contain duplicate values.
Unlike PRIMARY KEY, UNIQUE INDEX does not prohibit NULL values.
A UNIQUE INDEX can only be created when the column contains no duplicate values.
In table Car no duplicated plate numbers are allowed:
CREATE UNIQUE INDEX PlateNumberIndex ON Car (PlateNumber);
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482
CREATE INDEX
You can create multiple indexes on the same table. A single-column index is created on column Brand of table Car. A single-column index is created on column Color of the same table.
CREATE INDEX BrandsIndex ON Car (Brand); CREATE INDEX ColorsIndex ON Car (Color);
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483
CREATE INDEX
You can create a multi-column index. Here a two-column index is created:
CREATE INDEX BrandsColorsCombinationsIndex ON Car (Brand, Color);
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484
CREATE INDEX
By applying a multi-column UNIQUE INDEX, the relevant combination of columns can be defined as key.
In contrast to the PRIMARY KEY, UNIQUE INDEX does not prohibit NULL values in the key columns.
You can only create a multi-column UNIQUE INDEX, if the relevant column combination does not contain duplicates. Here, the value uniqueness of Country and PlateNumber combinations is enforced:
CREATE UNIQUE INDEX CountryPlateNumberCombiIndex ON Car_EU (Country, PlateNumber);
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485
Access Paths
DROP INDEX
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486
Access Paths
DROP INDEX Access_Path;
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487
DROP INDEX
You can drop an existing index. Only the access path (index) is deleted. The underlying data of the index is not deleted.
DROP INDEX does not syntactically distinguish between NON-UNIQUE and UNIQUE indexes.
DROP INDEX PlateNumberIndex;
© 2013
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488
Unit 10: Defining Data Access Summary
You should now be able to: •
Access tables in other schemas (assuming suitable permissions) • Control how other users can access data in your own schema • Explain when database indexes make sense even in HANA • Create and delete database indexes using SQL
© 2013
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489
Unit 11 Database Transactions
490
Unit 11: Database Transactions Learning Objectives
After completing this unit, you will be able to: • Explain what a database transaction is and why it is needed • Explain the acronym ACID
• Finish database transactions in HANA using SQL statements • Describe issues that arise if transactions are not mutually isolated • Understand how classical database systems synchronize concurrent transactions • Understand how HANA synchronizes concurrent transactions
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491
Database Transactions
•
Transaction (TA) = Sequence of associated database operations (SQL statements) for which the ACID requirements must be met.
•
All database operations always take place within transactions (in the extreme case, each operation has its own transaction).
• • • •
A C I D
© 2013
= = = =
Atomicity Consistency Isolation Durability
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492
Database Transactions
• Atomicity (A): A transaction is either executed completely or not at all. For Example: Transfer from Account1 to Account2
• Consistency (C): A transaction will bring the database from one consistent state to an(other) consistent state. During a transaction inconsistent states are possible, for example account transfers or marriage registration.
• Isolation (I): The database changes performed within a transaction shall only be visible to the outside after the completion of the transaction Important only for multi-user operations. Problematic if changes of canceled transactions can be seen.
• Durability (D): If a transaction is successfully completed (COMMIT), all changes from the transaction must permanently stay even in case of failures, or can be restored automatically. For example: Booking a deposit
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493
Database Transactions •
The ACID requirements must be ensured by the DBMS (not trivial).
•
The DBMS must provide mechanisms/automation for: Logging & Recovery for A + D Synchronization for I (such as locks) Integrity control for C
•
ACID requirements „cost“: With respect to the DBMS implementation effort With regard to performance at runtime
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494
Database Transactions
Which SQL statements can I use to start a transaction?
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495
Database Transactions
SAP HANA does not provide a SQL statement to explicitly start a transaction.
A transaction is implicitly started, if the previous transaction was canceled or successfully completed (or there is no previous transaction) AND one of the following SQL statements is executed SELECT INSERT UPDATE DELETE The transaction begins directly before the SQL statement, by which it is implicitly started.
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496
Database Transactions
Which SQL statements can I use to finish a transaction?
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497
Database Transactions
COMMIT
The currently running transaction should be completed successfully. The durability requirement must be ensured.
ROLLBACK
The currently running transaction should be canceled. The atomicity requirement must be ensured.
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498
Database Transactions
DELETE FROM Car; COMMIT; INSERT INTO Car(CarID) INSERT INTO Car(CarID) INSERT INTO Car(CarID) COMMIT; INSERT INTO Car(CarID) INSERT INTO Car(CarID) INSERT INTO Car(CarID) ROLLBACK; INSERT INTO Car(CarID) INSERT INTO Car(CarID) INSERT INTO Car(CarID) COMMIT; SELECT CarID FROM Car;
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VALUES ('F01'); VALUES ('F02'); VALUES ('F03'); VALUES ('F04'); VALUES ('F05'); VALUES ('F06'); VALUES ('F07'); VALUES ('F08'); VALUES ('F09');
CARID ----F01 F02 F03 F07 F08 F09
499
Database Transactions
DELETE FROM Car; COMMIT; INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES CREATE COLUMN TABLE T(S INT); INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES ROLLBACK; INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES COMMIT; SELECT CarID FROM Car;
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Certain SQL statements cause an implicit COMMIT ('F01'); ('F02'); ('F03'); ('F04'); ('F05'); ('F06'); ('F07'); ('F08'); ('F09');
CARID ----F01 F02 F03 F07 F08 F09
500
Multi-user Mode
Multi-user operations are absolutely necessary - but not without problems
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501
Multi-user Mode
Multi-user mode is absolutely necessary! •
If the transactions took place strictly in sequence one after another, this would be a huge waste of resources, since waiting times could not be used. • • •
Waiting for user input Waiting for disk access (for classical DBMS products) Waiting times caused by the application program
Multiple transaction must run time-shared (“interlocked”) •
When this happens in an uncontrolled manner (for example without synchronization), many problems can occur.
•
The problems can be divided into 5 different error classes (see following slides)
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502
Uncontrolled Multi-user Mode 1/5 Lost-update Problem The changes of a transaction are overwritten by another transaction and therefore lost.
Point in time
Transaction 1
t1
Read x
t2 t3
Read x x := x + 5,000; x := x – 100;
t4 t5 t6
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Transaction 2
Write x Write x
503
Uncontrolled Multi-user Mode 2/5 Inconsistent Analyses A transaction sees an apparent inconsistency, since it refers to two different consistent database states.
Point in time
Transaction 1
t1
Read marital status of Mandy (Result: “unmarried”)
t2
Change marital status of Mandy to “married”
t3
Change marital status of Thomas to “married”
t4
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Transaction 2
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Read marital status of Thomas (Result: “married”)
504
Uncontrolled Multi-user Mode 3/5 Phantom Problem (“malicious”, as it cannot be prevented by row-locking) A transaction sees an apparent inconsistency, since another transaction inserted a new row in the meantime. Point in time t1
Transaction 2
Read all SAP employees
t2
Insert Mr Z as a new SAP employee
t3
Register Mr Z as course participant
t4
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Transaction 1
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Read all course participants (Mr Z is a participant of an internal course, though he apparently seems to be no SAP employee!)
505
Uncontrolled Multi-user Mode 4/5 Non-repeatable read During the same transaction different results are obtained for repeated reading of the same facts. Point in time
Transaction 1
t1
Read salary of Mr A
t2 t3
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Transaction 2
Raise salary of Mr A Read salary of Mr A
506
Uncontrolled Multi-user Mode 5/5 Dependency on uncommitted changes (Dirty reads) A transaction sees and uses a state that officially never existed.
Point in time
Transaction 1
t1
Raise salary to € 10,000,000.-
t2
Read salary
t3
Donate € 5,000,000.-
t4
COMMIT;
t5
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Transaction 2
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ROLLBACK; (transaction canceled)
507
Synchronisation
To avoid the problems shown before, multi-user mode must be controlled or synrchonized – Synchronization = time based coordination of processes
A common option for synchronization is the use of locking – Locks are the most important (but not the only) way – “Multi-version Concurrency Control" can also be used – Alternatively, there is the “Optimistic Synchronization Method“
The goal of synchronization is the logical single-user mode with physical multi-user mode.
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508
Logical Single-user Mode
What does logical single-user mode with physical multi-user mode mean?
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509
Logical Single-user Mode
The user should have the impression of being the only active user of the database. This goal cannot be reached to 100%. If access to a (currently locked) object takes longer, the users will suspect that they are not alone. If the transaction is aborted by the DBMS to resolve a deadlock, the users will suspect that they are not alone.
Externally there should be the impression that the transactions are strictly executed in series (that means one after another - Serializability criterion) In reality: time-shared processing of several transactions – But the impression of serial processing should be created
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510
Serializability
What is the Serializability Criterion?
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511
Serializability
Serializability = There is a serial execution order of the transactions that has the same result – Result in terms of database state or database content
Serializability is therefore a correctness criterion for the "correct" execution of the synchronization (that means for the "right" control of the multi-user mode) Ensuring Serializability is the responsibility of the DBMS (and not of the user or of the application program) What Serializability means, you can best understand with reference to a concrete example … (see following slides).
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Transaction 1
UPDATE Official SET Overtime = Overtime + 2 WHERE PNr = 'P07'; UPDATE Official SET Overtime = Overtime + 2 WHERE PNr = 'P07'; COMMIT;
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513
Transaction 2
--Rounding to full ten UPDATE Official SET Overtime = ROUND(Overtime, -1) WHERE PNr = 'P07';
SELECT * FROM Contact; COMMIT;
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514
Transaction 3
UPDATE Official SET Overtime = Overtime * 4 WHERE PNr = 'P07'; UPDATE Official SET Overtime = Overtime / 2 WHERE PNr = 'P07'; COMMIT;
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515
Effect of the 3 Transactions
TA1 increases the overtime value by 4
1a) o = o + 2; 1b) o = o + 2; TA2 rounds the overtime value to full ten
2a) o = round(o); 2b) read access; TA3 doubles the overtime value
3a) o = o * 4; 3b) o = o / 2;
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Serial Execution
TA1 increases the overtime value by 4 TA2 rounds the overtime value to full ten TA3 doubles the overtime value
TA1 → TA2 → TA3
22 → 26 → 30 → 60
TA1 → TA3 → TA2
22 → 26 → 52 → 50
TA2 → TA1 → TA3
22 → 20 → 24 → 48
TA2 → TA3 → TA1
22 → 20 → 40 → 44
TA3 → TA1 → TA2
22 → 44 → 48 → 50
TA3 → TA2 → TA1
22 → 44 → 40 → 44
Initial value = 22
Correct result (according to Serializability): 44, 48, 50 und 60
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517
Incorrect Interleaving
Initial Value
22
1.
(1a)
o = o + 2;
24
2.
(2a)
o = round(o);
20
3.
(2b)
read access;
20
4.
(3a)
o = o * 4;
80
5.
(1b)
o = o + 2;
82
6.
(3b)
o = o / 2;
41
The interleaving shown above is incorrect, because the result (41) does not correspond to any value that can occur in serial execution (44, 48, 50 and 60)
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518
Correct Interleaving
Initial Value
22
1.
(3a)
o = o * 4;
88
2.
(2a)
o = round(o);
90
3.
(1a)
o = o + 2;
92
4.
(3b)
o = o / 2;
46
5.
(2b)
read access;
46
6.
(1b)
o = o + 2;
48
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The interleaving shown here is correct, because the result (48) corresponds to one of the values that can occur by serial execution (44, 48, 50 and 60).
519
Multi-user Mode
What are Isolation Levels?
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Isolation Levels
Isolation Levels define … •
Whether rows that are read, inserted, updated, or deleted by a transaction are available to other transactions running in parallel.
•
Whether read, insert, update, or delete activities of another transaction running in parallel have an impact on the current transaction.
•
Which problems of uncontrolled multi-user mode may occur and which problems must be prevented (by the DBMS)
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521
Isolation Levels
4 isolation levels are defined by the SQL Standard in such a way that each level defines which problems may occur and which problems must be prevented by the DBMS.
Isolation Level
Uncommitted Dependency (Dirty Reads)
Non-repeatable Reads
Phantom Problem
0
READ UNCOMMITTED
possible
possible
possible
1
READ COMMITTED
impossible
possible
possible
2
REPEATABLE READ
impossible
impossible
possible
3
SERIALIZABLE
impossible
impossible
impossible
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Multi-user Mode
Which SQL statement can I use to define the isolation level?
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523
Multi-user Mode
SET TRANSACTION ISOLATION LEVEL
A certain isolation level is set. You must set the isolation level before the transaction starts. The isolation level is only valid for a single (i.e. next to run) transaction. If different isolation levels are set before the transaction starts, only the last setting is decisive. The isolation level “SERIALIZABLE” guarantees serializability.
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524
Multi-user Mode
SET TRANSACTION ISOLATION LEVEL
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525
Isolation Levels
SET TRANSACTION ISOLATION LEVEL IsolationLevel;
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526
Isolation Levels
According to the SQL Standard isolation level SERIALIZABLE is the default. In SAP HANA the isolation level READ COMITTED is the default. It is not possible to set isolation level READ UNCOMMITED in SAP HANA. You can set the remaining three isolation levels as follows: SET TRANSACTION ISOLATION LEVEL READ COMMITTED; SET TRANSACTION ISOLATION LEVEL REPEATABLE READ; SET TRANSACTION ISOLATION LEVEL SERIALIZABLE;
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527
Synchronization
How do classical DBMS products synchronize the multi-user mode?
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528
Classical Synchronization
Classical DBMS products usually use locking for multi-user mode synchronization. Database objects (schemata, tables, rows) are locked prior to any access (and therefore cannot or only in a limited way be used by other transactions). Locks are set and released by the DBMS. User or applications do not need to care of the locks (automation of the DBMS).
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529
Synchronization
How does SAP HANA synchronize its multi-user mode?
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530
Synchronization in SAP HANA
The consistency of write access is ensured by X-locks The affected rows are locked “as usual”. The consistency of read access is ensured by “Multi Version Concurrency Control“ (MVCC) No S-locks are set when reading rows . By default, all SELECT statements of the same transaction see the same (possibly obsolete) consistent version of the database state (≈ isolation level REPEATABLE READ). This version contains all changes that were committed before the start of the transaction (together with all changes done by the transaction itself). This is called “transaction-level snapshot isolation”. Write access for other transactions running in parallel creates a new version of the database state (which is not visible to the transaction mentioned above.) Isolation level READ COMMITED corresponds to statement-level snapshot isolation
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Multi-Version Concurreny Control – An Example
Timeline of 5 transactions T1-T5 accessing same record „item D“
T2 T4
T5 starts after T3 is committed sees V2 When T4 ends (last transaction that still sees V1), V1 becomes obsolete
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V2 open V1 open
committed
read
read
insert
update
T5
item D versions
T3
T1 writes V1 of item D T3 starts and writes V2. V1 still remains visible for transactions that started before T3 ends. With „transaction-level“ snapshot isolation, T2 and T4 keep seeing V1 of item D (≈ repeatable read)
commit
Tx T1
read
commit
committed out-dated
532
Write Locks & Snapshot Isolation – Additional Remarks
Snapshot Isolation can lead to updates a transaction that started earlier doesn’t see Transaction-level snapshot isolation increases probability of conflicts on application level
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Transaction T1
write
write
read
re-read
Transaction T2
data record versions
Primary key & uniqueness constraints are checked considering all existing versions Deadlock situations can not be prevented
commit
V3 V2 V1
Committed Committed
533
Unit 11: Database Transactions Summary
You should now be able to: •
Explain what a database transaction is and why it is needed • Explain the acronym ACID • Finish database transactions in HANA using SQL statements • Describe issues that arise if transactions are not mutually isolated •
Understand how classical database systems synchronize concurrent transactions • Understand how HANA synchronizes concurrent transactions
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534
Summary
535
Summary
Motivation And Basic Concepts Reading Data From A Table Or View Aggregating Data
Reading Data From Multiple Tables Changing Data Stored in Tables Defining How Data Is Stored Using Views For Data Access Defining Data Access
Understanding NULL Values Database Transactions © 2013
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536
Open Questions
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537
Done!
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538
Meta Data
How can I access meta data?
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539
Accessing Meta Data
The DBMS provides metadata views (catalog views). This is a requirement of the SQL standard. You can use SQL to read the meta data.
Which tables belong to schema Schema1?
SELECT Table_Name FROM Sys.Tables WHERE Schema_Name = 'SCHEMA1';
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TABLE_NAME ---------CONTACT OFFICIAL OWNER STOLEN OWNER_EU CAR CAR_EU
540
Accessing Meta Data
Which columns are in table Car of schema Schema1?
SELECT FROM WHERE ORDER BY
Column_Name, Position, Data_Type_Name Sys.Table_Columns Schema_Name = 'SCHEMA1' AND Table_Name ='CAR' Position;
COLUMN_NAME ----------CARID PLATENUMBER BRAND COLOR HP OWNER
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POSITION -------1 2 3 4 5 6
DATA_TYPE_NAME -------------VARCHAR VARCHAR VARCHAR VARCHAR INTEGER VARCHAR
541
Accessing Meta Data
How many columns do the tables of schema Schema1 have?
SELECT FROM WHERE GROUP BY ORDER BY
Table_Name, COUNT(*) AS NumColumns Sys.Table_Columns Schema_Name = 'SCHEMA1' Table_Name 2 TABLE_NAME ---------CONTACT STOLEN OWNER OWNER OWNER_EU CAR CAR_EU
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NUMCOLUMNS ---------2 2 4 5 5 6 7 542
Accessing Meta Data
Which tables of schema Schema1 have the fewest columns?
SELECT FROM WHERE GROUP BY HAVING
TABLE_NAME ---------CONTACT STOLEN © 2013
Table_Name, COUNT(*) AS NumColumns Sys.Table_Columns Schema_Name = 'SCHEMA1' Table_Name COUNT(*) = (SELECT TOP 1 COUNT(*) FROM Sys.Table_Columns WHERE Schema_Name = 'SCHEMA1' GROUP BY Table_Name ORDER BY 1); NUMCOLUMNS ---------2 2
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543
Accessing Meta Data
Which tables of schema Schema1 have the most columns?
SELECT FROM WHERE GROUP BY HAVING
TABLE_NAME ---------CAR_EU
© 2013
Table_Name, COUNT(*) AS NumColumns Sys.Table_Columns Schema_Name = 'SCHEMA1' Table_Name COUNT(*) = (SELECT TOP 1 COUNT(*) FROM Sys.Table_Columns WHERE Schema_Name = 'SCHEMA1' GROUP BY Table_Name ORDER BY 1 DESC); NUMCOLUMNS ---------7
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544
Meta Data
Which meta data views are available?
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545
Accessing Meta Data
There are several meta data views …
Meta Data View
Information contained
Sys.Tables
Which tables are available?
Sys.Table_Columns
Which columns are in the tables?
Sys.Views
Which views are available?
Sys.View_Columns
Which columns are in the views?
Sys.Indexes
Which indexes are available?
Sys.Index_Columns
Which columns are in the indexes?
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546
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