Re: is there a way to deliver an array over column from a query window?

On Wed, Apr 24, 2013 at 2:44 AM, Rafał Pietrak <rafal(at)zorro(dot)isa-geek(dot)com> wrote:
> W dniu 03/24/2013 12:11 PM, Rafał Pietrak pisze:
>
> W dniu 03/24/2013 12:06 PM, Misa Simic pisze:
>
> maybe,
>
> SELECT DISTINCT issuer,amount, array_agg(REFERENCE) over (partition by
> invoice_nr) from invoices;
>
>
> RIGHT. Thenx. (and the first thing I did, I've read the doc on
> array_agg().... what stress makes from people :(
>
>
> Actually, I have a problem with that (which I haven't noticed earlier
> because the data I'm having, don't have to many "duplicates" that cause it).
> The problem is, that:
> --------------------------------------
> SELECT DISTINCT a, b, c, array_agg(k.d) OVER (PARTITION BY k.c ) FROM testy
> k where k.e <> 'email' and k.c='1035049' ;
> a | b | c | array_agg
> ------+----------+---------+-------------------------------
> 1035 | 10410053 | 1035049 | {9902031328529,5951948640868}
> ---------------------------------------
>
> is _almost_ fine. But I actually need to have control over the order in
> which the array gathered its values. So I try:
> ------------------------------------
> SELECT DISTINCT a, b, c, array_agg(k.d) OVER (PARTITION BY k.c ORDER BY k.e)
> FROM testy k where k.e <> 'email' and k.c='1035049' ;

you are aware of in-aggregate ordering (not completely sure if it
meets your use case?

select array_agg(v order by v desc) from generate_series(1,3) v;

also, 'distinct'
select array_agg(distinct v order by v desc) from (select
generate_series(1,3) v union all select generate_series(1,3)) q;

merlin