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SQL Get MIN row for each record

Posted on 2011-09-23
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Last Modified: 2012-06-27
Hello I have the following table data(posed in the CODE sec.  

wondering how to get the row data with the min date from each OPID.  If the 2nd Window date is less than 1st Window date then the 2nd Window should be read.    

get the min date goes something like...
Select distinct min DATE, OPID, Window
from TABLE

But the problem is if the 2nd window is the lesser date of the windows.
Accnt	OPID	DATE	Window 
34	1	9/22/11 12:32 PM	1st WINDOW
34	1	9/22/11 12:33 PM	2nd WINDOW
12	2	9/21/11 11:02 AM	1st WINDOW
12	2	9/21/11 11:01 AM	2nd Window
455	1	9/22/11 1:09pm	1st WINDOW

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Question by:TechMonster
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7 Comments
 
LVL 51

Expert Comment

by:HainKurt
Comment Utility
try:

select * from (
select *, row_number() over (partition by opid order by window) rn
from myTable ) x
where rn=1
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Accepted Solution

by:
HainKurt earned 500 total points
Comment Utility
oops, we should sue [DATE]

select * from (
select *, row_number() over (partition by opid order by [DATE]) rn
from myTable ) x
where rn=1
0
 

Author Comment

by:TechMonster
Comment Utility
i think that got it!  Have to test it out a few more times though...thanks.
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Author Comment

by:TechMonster
Comment Utility
Well it gets the min from all DATes..but I have to get the row by min dateTime for each day.
0
 
LVL 51

Expert Comment

by:HainKurt
Comment Utility
then you should partition by day & opid

easy way is to get date part

select floor(cast(GETDATE() as float))

select * from (
select *, row_number() over (partition by opid,  floor(cast([DATE] as float)) order by [DATE]) rn
from myTable ) x
where rn=1
0
 
LVL 51

Expert Comment

by:HainKurt
Comment Utility
or use CONVERT(varchar, getdate(), 101) -- to get rid of time part and group by date only like : 09/23/2011

select * from (
select *, row_number() over (partition by opid, CONVERT(varchar, getdate(), 101) order by [DATE]) rn
from myTable ) x
where rn=1
0
 

Author Comment

by:TechMonster
Comment Utility
Perfect..thanks again
0

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