|
Michael,
Thanks for your example. I've never seen the PARTITION BY before, but I
get the concept. The nice thing about your example is that I found it
useful to list only the 2nd duplicate, so I was able to change the last
line from "<= 2" to "= 2". (This helps me identify when the duplicates
started to occur, and it was quite illuminating for determining the root
cause.)
- Dan
On Mon, Jan 18, 2016 at 4:30 PM, Michael Schutte <mschutte369@xxxxxxxxx>
wrote:
you're right, order by needs added to the over(...) I missed that.mschutte369@xxxxxxxxx>
On Mon, Jan 18, 2016 at 4:12 PM, <rob@xxxxxxxxx> wrote:
Somewhere in there is a need for an order by PRDATER (process date).
But the analytical
row_number() over...
should be the light that clicks on for the OP.
Rob Berendt
From: Michael Schutte <mschutte369@xxxxxxxxx>
To: Midrange Systems Technical Discussion <midrange-l@xxxxxxxxxxxx
Date: 01/18/2016 04:05 PM
Subject: Re: Complex (for me) SQL question
Sent by: "MIDRANGE-L" <midrange-l-bounces@xxxxxxxxxxxx>
Add () to the row_number, remove group by from the over(...)
On Mon, Jan 18, 2016 at 3:53 PM, Michael Schutte <
--wrote:mschutte369@xxxxxxxxx>
Oops, didn't finish reading.
with sum1 as (
select PRPRDN from HPRDFA
group by PRPRDN
HAVING COUNT(1) > 1
), results as (
select HPRDFA.*, row_number over(partition by PRPRDN group by PRPRDN)
row_number_by_item
from HPRDFA
where PRPRDN in (select PRPRDN from sum1) )
select * from results
where row_number_by_item <= 2
On Mon, Jan 18, 2016 at 3:51 PM, Michael Schutte <
wrote:
Would this suffice?
select PRPRDN from HPRDFA
group by PRPRDN
HAVING COUNT(1) > 1
order by PRPRDN
Or
with sum1 as (
select PRPRDN from HPRDFA
group by PRPRDN
HAVING COUNT(1) > 1
)
select * from HPRDFA
where PRPRDN in (select PRPRDN from sum1)
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