SQL Bangla Tutorial .pdf
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Original filename: SQL Bangla Tutorial.pdf
Title: 6.SQL bangla tutorialsMicrosoft Word - 6.SQL bangla tutorials.docxMicrosoft Word - 6.SQL bangla tutorials.docx
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CREATE TABLE table_name( )
table_name e я* i 0 ( (।)u(,:
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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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+।):
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
)( p0 2l # - ,
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INSERT INTO test VALUES
‘WDAD’, ‘01195229300’)
(8,
'Rahman',
'Abidur',
' a_rahman@yahoo.com',
+
i = ' ( \ J +।)
1.INSERT INTO test(firstname, lastname) VALUES (‘Harun', ‘Rashid')
+d 3
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1.INSERT INTO test SET id = 9
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3.)( e' pi , , * e pi .e ei pi
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5. * 0u ) 0 )( i first name pi e )(
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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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я
SHOW TABLE ei sn 0 ( e ( )।
DESCRIBE TABLE ei sn ( s я+t ( )।) DESCRIBE test e0
0
(
TABLE ALTERATI ON
+=+= MYSQL e + 2i 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
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1.alter table table_name add column column_name data_type [NULL/not NULL]
e ( e )* +।e * o।
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select
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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
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:
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1.select JOB, SAL from EMPLOYEES where SAL > 1500
2 employees )( salary 1500 e u+ ( -# e (।
1.SELECT * FROM Persons
2.WHERE City='Sandnes'
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) sandnes =, - , u+
,।
SQL kk single quote e e - - = * - ।
6#(Numeric) = )।
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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 (
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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
pg6 e kt = ' e e - ( e a* 3 3
k -।i + SELECT statement e - SQL JOIN m! ( +।
e *d0 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” pi ,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 22
(।
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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