Yup, that did it. I don't know why I made it harder than it had to be.
Thank you.
Mike.
On Wednesday 15 August 2007 02:58:22 pm Fernando Hevia wrote:
> Try this:
>
> Select *
> from view v1
> where duration = (select max(duration) from view v2 where v2.phone_number =
> v1.phone_number)
>
> You could get more than one call listed for the same number if many calls
> match max(duration) for that number.
>
>
> -----Mensaje original-----
> De: pgsql-sql-owner@postgresql.org [mailto:pgsql-sql-owner@postgresql.org]
> En nombre de Mike Diehl
> Enviado el: Miércoles, 15 de Agosto de 2007 17:28
> Para: SQL Postgresql List
> Asunto: [SQL] Problem with phone list.
>
> Hi all.
>
> I've qot a problem I need to solve. I'm sure it's pretty simple; I just
> can't
> seem to get it, so here goes...
>
> I've got a table, actually a view that joins 3 tables, that contains a
> phone
>
> number, a unique id, and a call duration.
>
> The phone number has duplicates in it but the unique id is unique.
>
> I need to get a list of distinct phone numbers and the coorisponding
> largest
>
> call duration.
>
> I've got the idea that this should be a self-join on phone number where
> a.id<>b.id, but I just can't seem to get the max duration.
>
> Any hints would be much appreciated.
>
> Mike.
>
>
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--
Mike Diehl