SQL Bangla Tutorials

July 12, 2020 | Author: Anonymous | Category: Software Design, Data, Information Retrieval, Data Management Software, Computer Data
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SQL 

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SQL sn

  

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SQL Join Left Join

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e u u  u  - %& ( SQL Tut or i al i n Bangl a)

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SQL=Standard Query Language(uc  k  ek ess-cuell)    e6 a n  (query)я #

e e =k=  # 6* я.ei # 6* я ( )   s  +  (Manipulate) e6 a#   +  ) -MySQL,SQL Server,Oracle i# (। e 4thgeneration language. e  +!  G p g 6 # 6* я     s    +   ) loop,branch. e+& e u ,  MySQL Dat abase a0   1 2 3 %&   4।

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eu #  k   + e k  e  i  SQL  n  ,।

e u s5 u  ( SQL St at ement Tut or i al i n Bangl a)

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ALTER TABLE-  m (  , CREATE INDEX- INDEX  , DROP INDEX- INDEX  3 )  DML-Data Manipulation Language Query (a n ) e6 update SQL e DML a6= (  ,। SELECT-   -     UPDATE-  +(,  * () DELETE-   - -#  3 ( INSERT INTO-   -# J  

e u ,  >2 ? ( SQL Dat abase Cr eat i on Tut or i al i n Bangl a)

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MySQL e  я =  * e       ,।CREATE DATABASE ei sn ( e0     ,।XAMPP   3 ? -    e6 b uя    e eL   http://localhost/  en  ( e  e   (   Tools e a' phpMyadmin e k  =& SQL #  k  । e       (CREATE DATABASE designer) e6 go   k  #  e    , *

e    ( ( designer  e   3।

  k   e    e '   -  +  e - e0 k  # ,   i    sn 0 # ,   USE database_name( )  0 # ,   database_name e я *  i 0 ( ( ।)  B e ap я     3    sn DROP DATABASE database_name( )  0  3 database_name e я *  i 0 ( ( ।) e      3   ,    )   2  (undo) m , i     -  я0  , )  p я 0  3 ( ।

e u ,    >2 ? u  ( u  ( SQL Tabl e Cr eat i on Tut or i al i n Bangl a)

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  sn 0  m + CREATE TABLE table_name( )     table_name e я *  i 0 ( ( ।)u( , : designer  )  0  3   k  (  ( ( )e6 SQL #  k      n0  1.CREATE TABLE `designer`.`test` ( 2.`id` INT( 10 )NOT NULL AUTO_INCREMENT PRIMARYKEY , 3.`first name` VARCHAR( 15 ) NOT NULL , 4.`last name` VARCHAR( 15 ) NOT NULL , 5.`email` VARCHAR( 20 ) NOT NULL, 6.`course name` VARCHAR( 20 ) NOT NULL DEFAULT'WDAD', 7.`mobile` INT( 12 ) NOT NULL 8.)

e  GO   k  #  test  e   , )  designer database e a' ।   

u+   # # p-  i 0 (  ,c test  e   ।2  i (  ,c  id  e   i ) INT(10) a-# : +!  6# , (id  +!  6# i , )  o 1, o 20  o 230 )e6 100 = 6# , e+ 3 NOT NULL )    ei  0 2    )  । + AUTO_INCREMENT e    ,c + )( id (  ,

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e u i&C s5 u  ( SQL I nser t St at ement i n Bangl a)

  : яo

 яo  

eu (SQL) ( +  e  e '  (row) J   +  ।e  e  (row)e  , J   +  ।) : INSERT INTO test (id, first name, last name, email, course name,mobile) VALUES (8, 'Rahman', 'Abidur', ' [email protected]', ‘WDAD’, ‘01195229300’)

=  +d 3  J     ) (   e  u+  ) p0 2l я # #  -  ।( id, first name, last name e*  , e e0 2l) 2l )( p0 2l #  -   ,       J   , INSERT INTO test VALUES ‘WDAD’, ‘01195229300’)

(8,

'Rahman',

'Abidur',

' [email protected]',

+ i = '  ( \     J   +  ।) 1.INSERT INTO test(firstname, lastname) VALUES (‘Harun', ‘Rashid')

 +d 3 i  e  ( \   e   J   +  । 1.INSERT INTO test SET id = 9

e  a # # 2l*  2   2l #  - ।

e u pi ? u  ( SQL Pr i mar y Key Tut or i al i n Bangl a)

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1.e (  p0 entries  a #/ ( (uniquely)  h /= k  ,। 2.+ e  e '   p i   ,   ( +  ।) i  ei p i    i +   p0  (row)  a #/ ( (uniquely)  h । 3.)( e '   p i   ,    ,   *   e p i   .e  e i p i   - ।

5.p i   2  -  । 5.  * 0u   ) 0    )( i  first name  p i       e  )(  , я   -   ,    - o  - o e   -    ) o  m  3  , + first name  last name  p i     e  16   +)n   e PRIMARY KEY   )     o ) a # (unique) ,।eя # e  ) id number  PRIMARY KEY ,   ,  ( iя  id   o e , ।e   a +  )  (first name,last name e = = ,email e = = ii eL i# ( - )  J   ।

e u ,   ,+ B ( SQL Tabl e Del et e Show Tut or i al i n Bangl a)

  :

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SHOW TABLE ei sn 0 ( e     ( ) । DESCRIBE TABLE ei sn (  s  я +t ( ) ।) DESCRIBE test e0    0

(  TABLE ALTERATI ON

  + = + = MYSQL e +  2 i   a  a+= +  ) 1.alter table table_name add PRIMARY KEY (col_name,...)

2.alter table table_name ADD UNIQUE (col_name,...) 3.alter table table_name DROP PRIMARY KEY 4.alter table table_name ALTER [COLUMN] col_name {SET DEFAULT literal |DROP DEFAULT}

e d  +     ) *  ,  ,PRIMARY KEY    -   i# (  +  । 1.alter table table_name drop column column_name

e  table_name e я *     column_name e я *     ( oi  -  0  3 ( +  । 1.alter table table_name add column column_name data_type [NULL/not NULL]

e ( e     ) *  +  ।e   *  o।

e u  u  ( u  ( SQL Quer y Tut or i al i n Bangl a)

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  -    e ( SELECT sn 0 # ,  ,।) select

* from

test

e +  test  0 +    , я  2। *e я *  )(     (  , = ' oi  *  ( ।) SELECT firstname,lastname FROM

test

SELECT sn (    ( \ a6=o   e ( +  a-# : R  ( +   k + (

।) select JOB, SAL from EMPLOYEES Limit 10

' +  e   employees  e  3    , я   - i +  kt u+  n0 10 0   ( ।

e u WHERE Clause , 

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e (    n     ) ।) 1.select JOB, SAL from EMPLOYEES where SAL > 1500

2 employees  ) ( salary 1500 e u+  ( -# e (। 1.SELECT * FROM Persons 2.WHERE City='Sandnes'

' +    person   3 e + = '  ( -# )  sandnes =, -   , u+ 

 ,। SQL kk single quote e     e -  - = *       - । 6# (Numeric)      =      )  । e SELECT * FROM Persons WHERE FirstName='Tove e SELECT * FROM Persons WHERE FirstName=Tove 6#  kt e , e SELECT * FROM Persons WHERE Year=1965 e SELECT * FROM Persons WHERE Year='1965' WHERE clause e condition R  (    a+ *  # ,   ) 

R 

R 

AND e u( , : 1.SELECT * FROM Persons 2.WHERE FirstName='Tove' 3.AND LastName='Svendson'

OR e u( , : 1.SELECT * FROM Persons 2.WHERE FirstName='Tove' 3.OR FirstName='Ola'

AND,OR Operator e - # ,   condition   ) + ।) u( , : u( , 1.SELECT * FROM Persons WHERE 2.LastName='Svendson' 3.AND (FirstName='Tove' OR FirstName='Ola')

QUERY DI STI NCT e  3    p #  -  + ।    n0 ( ei  p eY  +  1.select distinct DEPTNO from EMPLOYEE

a-# : employee  - dept e m    ei +  n  a    -   i ei +  n m n e  a   -   i ? ei  n0 ( ei  p eY  m। QUERY ORDERBY 1.select ENAME, DEPTNO, HIREDATE from EMPLOYEE 2.order by DEPTNO [asc], HIREDATE desc

e  )  *  ( *  3  - Y(ascending)e6 Y - 3 (descending)  я ( +  । QUERY BETWEEN 1.select EMPNO, ENAME, SAL from EMP where SAL between 1500 and 2500 2.select ENAME from EMP where HIREDATE between ’02-APR-81’ and ’08-SEP-81

e (  ( \ e +  '#   a6= ( +  । 1.SELECT productCode, FROM products 2.WHERE buyPrice NOT BETWEEN 20 AND 100

e  AND i# ( # ,   QUERY  o (k  +  ।

e u  i& ( SQL Quer y I N Tut or i al i n Bangl a)

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1.select *

from DEPT where DEPTNO in (20,30)

   ( +  n ( o ) ( +  n m 20  30। 1.SELECT officeCode, city FROM offices 2.WHERE country IN ('USA','France') 3. 4. 5.SELECT officeCode, city FROM offices 6.WHERE country NOT IN ('USA','France')

e*  = s     ,  statement *  (i   a ) ।

a0n QUERY ( SQL Joi n Tut or i al i n Bangl a)

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a0n QUERY U+ ek =  )    SELECT statement ( e  -    e ( ) ।n    s

p g 6 e kt   = '  e   e  -  (   e  a *  3  3    k - । i + SELECT statement e  - SQL JOIN  m!   ( +  । e *d 0  m + 1.SELECT column_list 2.FROM table_1 3.[INNER |LEFT |RIGHT] table_2 ON conditions_2 4.[INNER |LEFT |RIGHT] table_3 ON conditions_3 5.… 6.WHERE conditions

JOIN keyword 0 SQL statement e  - # : , ( i   '  -      я # ) oi

*  '#  ( \    '#  m  u+  t  ,।*    '# e0 0  Keys(e*   Foreign key ) e  '# m ) k - । Primary Key ,c e  ( 3   \))   p0  (row)я # a #(unique).   “Persons” 0 ( P_Id

LastName

FirstName

Address

City

1

Hansen

Ola

Timoteivn 10

Sandnes

2

Svendson

Tove

Borgvn 23

Sandnes

3

Pettersen

Kari

Storgt 20

Stavanger

e  “Persons”  p i   ,c “P_Id”e    ,c ( 0  (row) “P_Id” o ei , ।ei “P_Id”i ( 0 row   (  )(o ( 0 row e  ei ,।e  + 0 ( O_Id

OrderNo

P_Id

1

77895

3

2

44678

3

3

22456

1

4

24562

1

5

34764

15

k#  “O_Id” 0 “Orders”  Primary Key e6 e  “P_Id” ( “Persons”  # k( p(=  ,3  (  i।ei  ( 0 '# “P_Id” 0  '# e0 m 3।

u( ,  *  '  JOIN 3  e ( JOIN,LEFT JOIN,RIGHT JOIN,FULL JOIN SQL I NNER JOI N INNER JOIN   -i      ) = (   u+  t ।) 1.SELECT column_name(s) 2.FROM table_name1 3.INNER JOIN table_name2 4.ON table_name1.column_name=table_name2.column_name

u( , : Persons table P_Id

LastName

FirstName

Address

City

1

Hansen

Ola

Timoteivn 10

Sandnes

2

Svendson

Tove

Borgvn 23

Sandnes

3

Pettersen

Kari

Storgt 20

Stavanger

Orders table O_Id

OrderNo

P_Id

1

77895

3

2

44678

3

3

22456

1

4

24562

1

5

34764

15

“P_Id”( ( 0 m ) k e - )(  # k(   ( i  ( order ) i ,    , INNER JOIN # ,      + । 1.SELECT Persons.LastName,Persons.FirstName,Orders.OrderNo 2.From Persons

3.INNER JOIN Orders 4.ON Persons.P_Id=Orders.P_Id 5.Order BY Persons.LastName

Result , e LastName

FirstName

OrderNo

Hansen

Ola

22456

Hansen

Ola

24562

Pettersen

Kari

77896

Pettersen

Kari

44678

INNER JOIN   (row) 2: + d  ) ( 0  '# +k e  - ।)( Persons table e  Orders table e  e  (row)  -   kt i  0 (  ।

e u  V я & u  ( SQL Lef t Joi n Tut or i al i n Bangl a)

  :

 я  

e   (table1) -     )(o a # (table_2,table_3...)  - e   - ।) 1.SELECT column_name(s) 2.FROM table_name1 3.LEFT JOIN table_name2 4.ON table_name1.column _name=table_name2.column_name

'  u+  - # k e6  ( k ( i       , 1.SELECT Persons.Lastname,FirstName,Orders.OrderNo 2.FROM Persons 3.LEFT JOIN Orders 4.ON Persons.P_Id=Orders.P_Id 5.ORDER BY Persons.LastName

Result ,     LastName

FirstName

OrderNo

Hansen

Ola

22456

Hansen

Ola

24562

Pettersen

Kari

77896

Pettersen

Kari

Svendson

Tove

44678

LEFT JOIN o  0 Left table(Persons) -   *    3 )(o Right table(Orders) e  - e

  i।

e u i я & u  ( SQL Ri ght Joi n Tut or i al i n Bangl a)

  :

 я  

e0 a # #  (table_2,table_3) -   )(o     - e   - ।) 1.SELECT Persons.Lastname,FirstName,Orders.OrderNo 2.FROM Persons 3.RIGHT JOIN Orders 4.ON Persons.P_Id=Orders.P_Id 5.ORDER BY Persons.LastName

e6 e 2 2     ( । LastName

FirstName

OrderNo

Hansen

Ola

22456

Hansen

Ola

24562

Pettersen

Kari

77896

Pettersen

Kari

44678 34764

SQL FULL JOI N e )(  # k k, e6 k  (  - # k ( i  ,     SELECT statement # ,   ,। 1.SELECT Persons.Lastname,FirstName,Orders.OrderNo 2.FROM Persons 3.FULL JOIN Orders 4.ON Persons.P_Id=Orders.P_Id 5.ORDER BY Persons.LastName

Result: LastName

FirstName

OrderNo

Hansen

Ola

22456

Hansen

Ola

24562

Pettersen

Kari

77896

Pettersen

Kari

44678

Svendson

Tove 34764

View more...

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