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# Finding dulplicate records

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I have the following #Temptable and would like to find out if there are duplicate records based on three fields

field 1, 2, 4

field1 | field2 | field3 | filed4 |Rule
1          1          2          2
1          1          2          3
1          1          2          4
1          1          2          4

If there are duplicate records then the 'rule' field gets a '1' else '0'

Results

field1 | field2 | field3 | filed4  |Rule
1          1             2        2         0
1          1             2        3         0
1          1             2        4         1
1          1             2        4         1

thanks,
lat
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Top Expert 2011

Commented:
update table
set rule = case when K > 1 then 1 else 0 end
from table as a inner join
(select field1,field2,field4 ,count(*) as K from table group by field1,field2,field4)  as b
on a.field1=b.field1 and a.field2=b.field2 and a.field4=b.field4

the above will work as long as none of field1,2,4 are null

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Commented:
If nulls are possible then just do this on lowfatspread's query:

IsNull(a.field1, 0) =IsNull(b.field1,0) and IsNull(a.field2, 0)=IsNull(b.field2,0) and IsNull(a.field4, 0) = IsNull( b.field4,0)
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Top Expert 2011

Commented:
thanks namasi,
i think it s time for me to get some sleep
;-)
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Commented:

Isn't not too early to get some sleep.?

regards-
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Top Expert 2011

Commented:
2am is kind of late for me...
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Commented:
I live in Canada. It is only 9 PM here.. Have a good night sleep...
Senior DBA
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Most Valuable Expert 2018
Distinguished Expert 2019

Commented:
Sometimes the EXISTS method will perform better (probably peforms better less often overall than Lowfat's method, but it's still a good technique to know :-) ):

UPDATE #temptable
SET [rule] = CASE WHEN EXISTS(
SELECT 1
FROM #temptable temp2
WHERE #temptable.field1 = temp2.field1
AND #temptable.field2 = temp2.field2
AND #temptable.field4 = temp2.field4) THEN 1 ELSE 0 END

Same as above for dealing with NULL(s), if any.

Commented:
hello  Lowfatspread, i think your query works but i';m running it on about 5k records, and so it may take a little bit of time to verify if the results are correct.  can you explain how the syntax works, i'm new to sql this and would like to learn what it means.

thanks,
lat
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Top Expert 2011
Commented:
This one is on us!
(Get your first solution completely free - no credit card required)

Commented:
Thanks for all of your help!

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