Re: sql help
- From: Igor Neyman <igor.neyman@xxxxxxxxx>
- To: JSweetser@xxxxxxxx
- Date: Wed, 21 Dec 2011 13:55:19 -0500
Trying it the second time (first time didn't go through because of
"over-quoting").
Is this what you are looking for:
WITH aggr AS (SELECT message, min(created_time) min_time FROM joe GROUP BY
message)
SELECT j.*
FROM joe j, aggr
WHERE j.message = aggr.message
AND j.created_time = aggr.min_time;
?
Regards,
Igor Neyman
On Wed, Dec 21, 2011 at 1:38 PM, Sweetser, Joe <JSweetser@xxxxxxxx> wrote:
> I don't think this should be as hard as I am making it. I want to select
> the earlier of all records that have the same message. There is a bit of
> extraneous data in here but I'm lazy and it was easier to leave it in so my
> apologies for that. I hope this formats ok.
> CREATE TABLE "JOE" ( "ID_ACTIVITY" NUMBER(10),
> "CREATED_USERNAME" VARCHAR2(100),
> "OWNER_USERID" VARCHAR2(100),
> "CREATED_TIME" TIMESTAMP(6),
> "CATAGORY" VARCHAR2(8),
> "MESSAGE" VARCHAR2(2000));
>
>
> So, I am looking for a statement that would return rows 1, 3, 5, and 7. I
> have tried various things with min(created_time) but that is too limiting
> since I only get one row (duh).
>
> Any/all pointers/tips welcome.
>
> thanks,
> -joe
>
>
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