How i can write query return the douplicate record with deffrint IDs ?! [modified]

[AJ Hoge]

Dear d@nish , Here i have different case , becuse i have two douplicate record with different ID , and when i use max() or top() function i will be ignor one of the ID ,and this couse problem for me to link to other tables. so i need to display all the two records with there IDs. ould you please help me to do that ?! Thanks

[dan sh]

If you will read Shameel's reply to your earlier post, you should be able to figure it out. Anyways, a general query would be:

select [colNames]
from tableName
group by [colNames]
having count(*) > 1

50-50-90 rule: Anytime I have a 50-50 chance of getting something right, there's a 90% probability I'll get it wrong...!!

[AJ Hoge]

you are right and i know that ! but if i write this :

select [ID,colNames,Phone,...]from tableNamegroup by [colNames]having count(*) > 1

i will have 0 douplicate record !! , why ? because IDs for each douplicted records are different !! and if use all columns without IDs ,i will got douplicted record without IDs. please see the example in my question , i don't need to ignore any IDs. could you please help me to find way? Thanks .

[Luc Pattyn]

All you need is a variation on SQL code shown by others. Try this:

SELECT ID, otherfields FROM table
GROUP BY otherfields
HAVING COUNT(*) > 1

i.e. don't use MAX(ID) or TOP(ID) and include all fields ("otherfields") that are relevant for identity checking :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

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I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

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[puri keemti]

Use CTE feature of SQL server 2005..... Example:

;WITH SalesCTE(Product, SaleDate, SalePrice, Ranking)

AS

(

SELECT

Product, SaleDate, SalePrice,

Ranking = DENSE_RANK() OVER(PARTITION BY Product, SaleDate, SalePrice ORDER BY NEWID() ASC)

FROM SalesHistory

)

DELETE FROM SalesCTE

WHERE Ranking > 1

[dan sh]

As Luc mentioned, you don't need to group by ID. You will get the duplicate records then.

50-50-90 rule: Anytime I have a 50-50 chance of getting something right, there's a 90% probability I'll get it wrong...!!

[AJ Hoge]

Hi Luc ,you are right, I Wrote that what you say , but i got error that " you try to excute a query that does not include the specified expression 'ID' as part of an aggregate function" Thanks

[AJ Hoge]

Dear d@nish , see my replay to LUC. Thanks

[AJ Hoge]

Dear puri keemti, i do not use SQL server , i use sql in access 2007.and i dont need to delet the record that douplicated , i need to display it as it is ,with there different IDS . if it is not clear please see question and my replies. could you help me ? thanks for you brother. :)

[Blue_Boy]

Hi, This query works for SQL Server2005,but try to convert for Access2007. A)

CREATE TABLE #tmpData
(
id INT IDENTITY(1,1),
custname VARCHAR(MAX)
)

INSERT INTO #tmpData
SELECT m.Customername
FROM mytable1 m
GROUP BY m.Customername,m.phonenumber,m.city,m.street HAVING COUNT(m.Customername)>1

DECLARE @i AS INT
DECLARE @query AS nvarCHAR(MAX)

SET @i=0
SET @query = 'select * from mytable1 where customername in ('

WHILE(@i<(SELECT COUNT(*) FROM #tmpData))
BEGIN
SET @i=@i+1
SET @query=@query+(SELECT char(39)+custname + char(39) FROM #tmpData WHERE id=@i)+','

END
SET @query=SUBSTRING(@query,1,LEN(@query)-1)
SET @query=@query+')'

EXEC( @query)
DROP TABLE #tmpData

I just played around and found too way to work for Access (I tested it on Access 2003 and not for 2007 version,but i think it should work too for Access2007 version) B)

select a.id,a.Customername , a.phonenumber, a.city, a.street
from mytable1 a
inner join
(select Customername , phonenumber, city, street
from mytable1
group by customername,phonenumber,city,street
having count(*) > 1) b
on a.Customername = b.Customername

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modified on Monday, December 7, 2009 9:32 AM

[Niladri_Biswas]

Try this Will work for Sql Server(Any version), MySql,Oracle, MSAccess

select * from myTable where customername in (
select customername
from myTable
group by customername,phonenum,city,street
having(count(customername)>1 and count(phonenum)>1 and count(city)>1 and count(street)>1))

Output:

id customername phonenum city street
19 jon 555 NK st.5
37 jon 555 NK st.5

Hope this helps :)

Niladri Biswas