deleting duplicate rows? - sql

Question

i want to delete duplicate rows from my table on the basis of category ID, but don't want to delete all, i want to left one rows if there are more than one row with the same category ID.
this is my query i am making i need to change it.
delete from twinhead_tblcategory where categoryid in (select categoryid from twinhead_tblcategory group by categoryid having count(categoryid) > 1 )

Answer 1

For SQL Server you can do it:
WITH MyTableCTE (CategoryId, RowNumber)
AS
(
SELECT CategoryId, ROW_NUMBER() OVER (ORDER BY CategoryId) AS 'RowNumber'
FROM MyTable
)
Delete From MyTableCTE Where RowNumber > 1

Answer 2

Do a select distinct into a new table, delete the old one and rename the new one into old table name.

Answer 3

If your rows have a distinct id column, then this should work:
DELETE t1 FROM your_table t1, your_table t2
WHERE t1.column1 = t2.column1 AND t1.column2 = t2.column2
AND ... /* check equality of all relevant columns */
AND t1.id < t2.id

Answer 4

Check here for sql server - http://support.microsoft.com/kb/139444 - that should get you started.

Answer 5

This is probably heavy-handed but perhaps you could select distinct * into a temp table, then truncate the table, then insert into the table the contents of the temp table. Foreign key constraints may prevent this, though.

Answer 6

For SqlServer, you could use a cursor to loop through all items, ordered by that categoryID.
Is the current ID the same as the previous one? Then delete it, see example C of this article.
Else remember the ID for the next round.

Answer 7

You have several way for delete duplicate rows.
for my solutions , first consider this table for example
CREATE TABLE #Employee
(
ID INT,
FIRST_NAME NVARCHAR(100),
LAST_NAME NVARCHAR(300)
)
INSERT INTO #Employee VALUES ( 1, 'Vahid', 'Nasiri' );
INSERT INTO #Employee VALUES ( 2, 'name1', 'lname1' );
INSERT INTO #Employee VALUES ( 3, 'name2', 'lname2' );
INSERT INTO #Employee VALUES ( 2, 'name1', 'lname1' );
INSERT INTO #Employee VALUES ( 3, 'name2', 'lname2' );
INSERT INTO #Employee VALUES ( 4, 'name3', 'lname3' );
First solution : Use another table for duplicate rows.
SELECT DISTINCT *
FROM #Employee
SELECT * INTO #DuplicateEmployee
FROM #Employee
INSERT #DuplicateEmployee
SELECT DISTINCT *
FROM #Employee
BEGIN TRAN
DELETE #Employee
INSERT #Employee
SELECT *
FROM #DuplicateEmployee
COMMIT TRAN
DROP TABLE #DuplicateEmployee
SELECT DISTINCT *
FROM #Employee
Second solution :
SELECT DISTINCT * FROM #Employee
SELECT * INTO #DuplicateEmployee FROM #Employee
INSERT #DuplicateEmployee
SELECT ID,
FIRST_NAME,
LAST_NAME
FROM #Employee
GROUP BY
ID,FIRST_NAME,LAST_NAME
HAVING COUNT(*) > 1
BEGIN TRAN
DELETE #Employee
FROM #DuplicateEmployee
WHERE #Employee.ID = #DuplicateEmployee.ID
AND #Employee.FIRST_NAME = #DuplicateEmployee.FIRST_NAME
AND #Employee.LAST_NAME = #DuplicateEmployee.LAST_NAME
INSERT #Employee
SELECT *
FROM #DuplicateEmployee
COMMIT TRAN
DROP TABLE #DuplicateEmployee
SELECT DISTINCT * FROM #Employee
teared solution : use rowcount
SELECT DISTINCT *
FROM #Employee
SET ROWCOUNT 1
SELECT 1
WHILE ##rowcount > 0
DELETE #Employee
WHERE 1 < (
SELECT COUNT(*)
FROM #Employee a2
WHERE #Employee.ID = a2.ID
AND #Employee.FIRST_NAME = a2.FIRST_NAME
AND #Employee.LAST_NAME = a2.LAST_NAME
)
SET ROWCOUNT 0
SELECT DISTINCT *
FROM #Employee
Fourth solution : use Analytical Functions
SELECT DISTINCT *
FROM #Employee;
WITH #DeleteEmployee AS (
SELECT ROW_NUMBER()
OVER(PARTITION BY ID, First_Name, Last_Name ORDER BY ID) AS
RNUM
FROM #Employee
)
DELETE
FROM #DeleteEmployee
WHERE RNUM > 1
SELECT DISTINCT *
FROM #Employee
Fifth solution : Use identity field
SELECT DISTINCT *
FROM #Employee;
ALTER TABLE #Employee ADD UNIQ_ID INT IDENTITY(1, 1)
DELETE
FROM #Employee
WHERE UNIQ_ID < (
SELECT MAX(UNIQ_ID)
FROM #Employee a2
WHERE #Employee.ID = a2.ID
AND #Employee.FIRST_NAME = a2.FIRST_NAME
AND #Employee.LAST_NAME = a2.LAST_NAME
)
ALTER TABLE #Employee DROP COLUMN UNIQ_ID
SELECT DISTINCT *
FROM #Employee
and end of all solution use this command
DROP TABLE #Employee
Source of my answer is this site