Обсуждение: Select groupping by one column

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Select groupping by one column

От
Josué Maldonado
Дата:
Hello list,

I have a table called pedh that looks like this:

REFNO    FECHA        OCNUM    PVD
0199    10/12/2003    5224    632
0199    10/12/2003    5224    632
1264    10/18/2003    8991    210
1264    10/18/2003    8991    210
1264    10/18/2003    8991    210
9093    10/20/2003    6895    520
9093    10/20/2003    6895    520

I need to select one record groupped by column refno, I tried using this
code:

select refno, max(fecha), max(ocnum), max(pvd) from pedh

It seems to work fine, but is there another way to get the same result
avoing use of aggregate functions?


Thanks in advance




--
Josué Maldonado



Re: Select groupping by one column

От
Bruno Wolff III
Дата:
On Thu, Dec 18, 2003 at 15:09:52 -0800,
  Josué Maldonado <josue@lamundial.hn> wrote:
> Hello list,
>
> I have a table called pedh that looks like this:
>
> REFNO    FECHA        OCNUM    PVD
> 0199    10/12/2003    5224    632
> 0199    10/12/2003    5224    632
> 1264    10/18/2003    8991    210
> 1264    10/18/2003    8991    210
> 1264    10/18/2003    8991    210
> 9093    10/20/2003    6895    520
> 9093    10/20/2003    6895    520
>
> I need to select one record groupped by column refno, I tried using this
> code:
>
> select refno, max(fecha), max(ocnum), max(pvd) from pedh

I don't see how that could work. Maybe you left off group by. But if you
did the aggreates could come from different rows, which might not be what
you want.

>
> It seems to work fine, but is there another way to get the same result
> avoing use of aggregate functions?

You might be able to use distinct on instead of aggregates.

Re: Select groupping by one column

От
Josué Maldonado
Дата:
Hello Bruno,

Bruno Wolff III wrote:
>>
>>select refno, max(fecha), max(ocnum), max(pvd) from pedh
>
>
> I don't see how that could work. Maybe you left off group by. But if you
> did the aggreates could come from different rows, which might not be what
> you want.

I missed the group by 1, I'm sorry.

>
>
>>It seems to work fine, but is there another way to get the same result
>>avoing use of aggregate functions?
>
>
> You might be able to use distinct on instead of aggregates.

Thanks