It helps me alot, the answer to your cuestion is: “depens of the amount of data” in my case I compare two tables with some hundreds of million of records…. if is your case, JOIN must be your choice.

Any one Please help me to simplify this query and i joined more than 6 tables it will take more time to execute the process. Anyone Please help me to favour……..

Thanks in advance

Last line should be:
(T1.col2 = 1) AND (T1.col3 = 2) AND (T2.col1 IS NULL)

CODE:
SELECT col1, col2, col3
FROM dbo.myTable AS T1
WHERE (col2 = 2) AND (col3 = 1) OR
(col2 = 1) AND (col3 = 2)
AND (col1 NOT IN (SELECT col1
FROM dbo.myTable
WHERE (col2 = 2) AND (col3 = 1)))

Is the same as…

CODE:
SELECT col1, col2, col3
FROM dbo.myTable AS T1 LEFT OUTER JOIN
dbo.myTable AS T2 ON T1.col1 = T2.col1
WHERE (T1.col2 = 2) AND (T1.col3 = 3) OR
(T1.col2 = 1) AND (T2.col3 = 2) AND (T2.col1 IS NULL)

The reason I wanted to find a way to convert this subquery into a join was that I am trying to create this as an indexed view. Figured I’d post this response in hopes that it may help someone else in the future. :)

CODE:
SELECT col1, col2, col3
FROM dbo.myTable AS T1
WHERE (col2 = 2) AND (col3 = 1) OR
(col2 = 1) AND (col3 = 2)
AND (col1 NOT IN (SELECT col1
FROM dbo.myTable
WHERE (col2 = 2) AND (col3 = 1)))

‘Not Show result
Select ’02′,ccd01emp,ccd01ano,ccd01mes,ccd01subd,
ccd01numer,ccd01ord,ccd01cta,ccd01deb,ccd01hab,ccd01con
from ccd left outer Join ccm01cta
On ccd01cta = ccm01cta
Where
ccd01emp=’01′
And ccm01emp=’01′
And ccd01ano=’2011′
And ccm01aa=’2011′
And ccd01mes=’05′
And isnull(ccm01cta,”)=”
‘

This query returned 0 records.

SELECT DISTINCT(A.CSN_ID) FROM TRG_LINKED_ENCS A
WHERE A.CSN_ID not in (SELECT DISTINCT(CSN_ID) FROM
TPL_TXDAYS )

This query retured 8700 records.

SELECT A.CSN_ID FROM TRG_LINKED_ENCS A
LEFT JOIN TPL_TXDAYS B ON
A.CSN_ID =B.CSN_ID
WHERE B.CSN_ID IS NULL

I- CREACIÓN TABLAS:
CREATE TABLE TB_DEMO_A( ID NUMBER, DESCRIPCION VARCHAR2( 10 ) );
CREATE TABLE TB_DEMO_B( ID NUMBER, DESCRIPCION VARCHAR2( 10 ) );

II- CREACIÓN INSERTS:
INSERT INTO TB_DEMO_A VALUES( 1, ‘PRIMERO’ );
INSERT INTO TB_DEMO_A VALUES( 2, ‘SEGUNDO’ );
INSERT INTO TB_DEMO_A VALUES( 3, ‘TERCERO’ );
INSERT INTO TB_DEMO_A VALUES( 4, ‘CUARTO’ );

INSERT INTO TB_DEMO_B VALUES( 1, ‘PRIMERO’ );
INSERT INTO TB_DEMO_B VALUES( 2, ‘SEGUNDO’ );
INSERT INTO TB_DEMO_B VALUES( 5, ‘QUINTO’ );
INSERT INTO TB_DEMO_B VALUES( 6, ‘SEXTO’ );

III- SELECTS:
SELECT a.* FROM TB_DEMO_A a;
SELECT b.* FROM TB_DEMO_B b;

IV.- AMARRES X CASO:

4.1.- “CONJUNTO A” INTERSECCION “CONJUNTO B”:

4.1.1.- FORMA #1:
SELECT a.*, b.*
FROM TB_DEMO_A a,
TB_DEMO_B b
WHERE a.ID = b.ID;

4.1.1.- FORMA #2:
SELECT a.*, b.*
FROM TB_DEMO_A a
INNER JOIN TB_DEMO_B b
ON a.ID = b.ID;

4.2.- (“CONJUNTO A” INTERSECCION “CONJUNTO B”) + ( “CONJUNTO A” MENOS CONJUNTO B” ):

4.2.1.- FORMA #1:
SELECT a.*, b.*
FROM TB_DEMO_A a,
TB_DEMO_B b
WHERE a.ID = b.ID(+);

4.2.2.- FORMA #2:
SELECT a.*, b.*
FROM TB_DEMO_A a
LEFT JOIN TB_DEMO_B b
ON a.ID = b.ID;

4.3.- (“CONJUNTO A” INTERSECCION “CONJUNTO B”) + ( “CONJUNTO B” MENOS CONJUNTO A” ):

4.3.1.- FORMA #1:
SELECT a.*, b.*
FROM TB_DEMO_A a,
TB_DEMO_B b
WHERE a.ID(+) = b.ID;

4.3.2.- FORMA #2:
SELECT a.*, b.*
FROM TB_DEMO_A a
RIGHT JOIN TB_DEMO_B b
ON a.ID = b.ID;

4.4.- ( “CONJUNTO A” MENOS CONJUNTO B” ):

4.4.1.- FORMA #1:
SELECT a.*, b.*
FROM TB_DEMO_A a,
TB_DEMO_B b
WHERE a.ID = b.ID(+) AND
b.ID IS NULL;

4.4.2.- FORMA #2:
SELECT a.*, b.*
FROM TB_DEMO_A a
LEFT JOIN TB_DEMO_B b
ON a.ID = b.ID
WHERE b.ID IS NULL

4.4.3.- FORMA #3:
SELECT a.*
FROM TB_DEMO_A a
WHERE a.ID
NOT IN ( SELECT b.ID
FROM TB_DEMO_B b
);

4.4.4.- FORMA #4:
SELECT a.*
FROM TB_DEMO_A a
WHERE NOT EXISTS( SELECT b.ID
FROM TB_DEMO_B b
WHERE a.ID = b.ID
);

4.5.- ( “CONJUNTO B” MENOS CONJUNTO A” ):

4.5.1.- FORMA #1:
SELECT a.*, b.*
FROM TB_DEMO_A a,
TB_DEMO_B b
WHERE a.ID(+) = b.ID AND
a.ID IS NULL;

4.5.2.- FORMA #2:
SELECT a.*, b.*
FROM TB_DEMO_A a
LEFT JOIN TB_DEMO_B b
ON a.ID = b.ID
WHERE a.ID IS NULL

4.5.3.- FORMA #3:
SELECT b.*
FROM TB_DEMO_B b
WHERE b.ID
NOT IN ( SELECT a.ID
FROM TB_DEMO_A a
);

4.5.4.- FORMA #4:
SELECT b.*
FROM TB_DEMO_B b
WHERE NOT EXISTS( SELECT a.ID
FROM TB_DEMO_A a
WHERE b.ID = a.ID
);

Diferents Style to Make the same.

e..g
select mobileno from table1 where (boxno,fileno) not in(select boxno,fileno from tabel2)

Most of the time I heard Join is better than subqueries

CREATE FUNCTION fn_get_days_forward
(
@FirstDay datetime
)
RETURNS @dates TABLE
(
dt DATETIME
)
AS
BEGIN
DECLARE @nextDay DATETIME
SELECT TOP 1 @nextDay = DATEADD(DAY, -1, dt) FROM dates WHERE dt > @FirstDay ORDER BY dt ASC

SET @FirstDay = DATEADD(DAY, 1, @FirstDay)
WHILE @FirstDay < @nextDay
BEGIN
INSERT INTO @dates SELECT @FirstDay

SET @FirstDay = DATEADD(DAY, 1, @FirstDay)
END

RETURN
END
GO

SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, days.dt))
FROM dates CROSS APPLY fn_get_days_forward(dt) AS days

Consider these as sample dates, now my output should be
03/01/09
09/01/09
10/01/09
11/01/09

Either give sample data or give us more details. When you say 24 Hours, is this 24 hours between two successful dates ? Please help us to help you by providing more information.

~ IM.

To put it in simple, i want to know the transaction of a particular inventory for which there is no transaction on that particular date, i want to know all the dates of year(with 24 hrs gap)

Regards,
Pradip

Thanks