Re: Subsorting GROUP BY data - Mailing list pgsql-sql

From Johnson, Michael L.
Subject Re: Subsorting GROUP BY data
Date
Msg-id D9865BC8CAFD9547AF959660DE822764015FBCA7@XMBIL133.northgrum.com
Whole thread Raw
In response to Re: Subsorting GROUP BY data  ("Fernando Hevia" <fhevia@ip-tel.com.ar>)
List pgsql-sql
Thanks! That's perfect, because now I don't need the FIRST/LAST
aggregate functions!

Mike

-----Original Message-----
From: pgsql-sql-owner@postgresql.org
[mailto:pgsql-sql-owner@postgresql.org] On Behalf Of Fernando Hevia
Sent: Monday, November 10, 2008 10:30 AM
To: Johnson, Michael L.; pgsql-sql@postgresql.org
Subject: Re: [SQL] Subsorting GROUP BY data


> -----Mensaje original-----
> De: pgsql-sql-owner@postgresql.org
> [mailto:pgsql-sql-owner@postgresql.org] En nombre de Johnson, Michael
> L.
> Enviado el: Lunes, 10 de Noviembre de 2008 12:57
> Para: pgsql-sql@postgresql.org
> Asunto: [SQL] Subsorting GROUP BY data
>
> Given the following table:
>
> ID  |  Cat  |  Num
> ----|-------|------
> Z   |   A   |   0
> Y   |   A   |   1
> X   |   A   |   2
> W   |   B   |   0
> V   |   B   |   1
> U   |   B   |   2
> T   |   C   |   0
> S   |   C   |   1
> R   |   C   |   2
>
> I want to do this:  Group the items by the cat field.  Then select the

> ID where the num is the highest in the group; so it should return
> something like:
>
> Cat  |  ID  |  Num
> -----|------|------
>   A  |  X   |   2
>   B  |  U   |   2
>   C  |  R   |   2
>
>
> Using SQL like this, I can get the category and the highest # in the
> category:
>
> SELECT cat, MAX(num) FROM my_table GROUP_BY cat;
>
> But if I add the "id" column, of course it doesn't work, since it's
> not in an aggregate function or in the GROUP_BY clause.  So I found a
> post at
> http://archives.postgresql.org/pgsql-hackers/2006-03/msg01324.php
> which describes how to add a "FIRST" and "LAST" aggregate function to
> PGSQL.  However, first and last don't seem to help unless you are able

> to "subsort" the grouping by the # (ie, group by cat, then subsort on
> num, and select the "last"
> one of the group).
>

I wonder if this suites you:

SELECT sub.cat, t.id, sub.Num FROM my_table t, ( SELECT cat, MAX(num) as Num FROM my_table GROUP_BY
cat
) subWHERE t.cat = sub.cat AND t.Num = sub.Num
ORDER BY t.cat;


Regards,
Fernando.



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