Re: Subsorting GROUP BY data

Thanks! That's perfect, because now I don't need the FIRST/LAST
aggregate functions!

Mike

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

> -----Mensaje original-----
> De: pgsql-sql-owner(at)postgresql(dot)org
> [mailto:pgsql-sql-owner(at)postgresql(dot)org] En nombre de Johnson, Michael
> L.
> Enviado el: Lunes, 10 de Noviembre de 2008 12:57
> Para: pgsql-sql(at)postgresql(dot)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
) sub
WHERE t.cat = sub.cat AND t.Num = sub.Num
ORDER BY t.cat;

Regards,
Fernando.