SQL Bangla Tutorial (PDF)




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Title: 6.SQL bangla tutorialsMicrosoft Word - 6.SQL bangla tutorials.docxMicrosoft Word - 6.SQL bangla tutorials.docx
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DML-Data Manipulation Language
Query (a n ) e6 update SQL e DML a6= (  ,।
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database_name e я *  i 0 ( ( ।)
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DROP DATABASE database_name( )  0  3
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u  ( SQL Tabl e Cr eat i on Tut or i al
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CREATE TABLE table_name( )    
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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.)

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p-  i 0 (  ,c test  e   ।2  i (  ,c  id  e  
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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', ' a_rahman@yahoo.com', ‘WDAD’, ‘01195229300’)

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

(8,

'Rahman',

'Abidur',

' a_rahman@yahoo.com',

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

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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)
  :
 я  

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}

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1.alter table table_name drop column column_name

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( +  ।
1.alter table table_name add column column_name data_type [NULL/not NULL]

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select

* from

test

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*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 , 
  :
 я  

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)
  :
 я  

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)
  :
 я  

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






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