--- Michael Glaesemann wrote:
> SELECT DISTINCT ON (phone_number)
> phone_number, call_duration, id
> FROM calls
> ORDER BY phone_number
> , call_duration DESC;
Wasn't acquainted with "DISTINCT ON (column)".
I found it to be many times faster than other suggestions using JOIN.
Cheers,
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
On Aug 15, 2007, at 15:28 , Mike Diehl wrote:
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
co
--- Mike Diehl <[EMAIL PROTECTED]> wrote:
> 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
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: [EMAIL PROTECTED] [mailto:[E
On 8/15/07, Mike Diehl <[EMAIL PROTECTED]> wrote:
> Any hints would be much appreciated.
DDL + sample data, please...
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