Re: [SQL] SubSelect as a column

Поиск
Список
Период
Сортировка
От Lars
Тема Re: [SQL] SubSelect as a column
Дата
Msg-id Pine.BSF.4.10.10002081103040.56244-100000@maximillion.sscsinc.com
обсуждение исходный текст
Ответ на Re: [SQL] SubSelect as a column  (Patrick JACQUOT <patrick.jacquot@anpe.fr>)
Список pgsql-sql
Thanks, I will give that a try. Hopefully the NOT EXISTS will not be too
slow. It is my understanding, I could be very wrong, that NOT EXISTS will
not take advantage of my indexes.

On another note, any idea when PostgreSQL 7 will be released? It sure
would be nice to have the subselect as a colunm feature.

Thanks,

-Lars

On Tue, 8 Feb 2000, Patrick JACQUOT wrote:

> Tom Lane wrote:
> 
> > Lars <lars@sscsinc.com> writes:
> > > I am trying to use a subselect as a column name, but it appears as if this
> > > is not supported in Postgresql. Here is the query:
> >
> > > SELECT u.idnum, u.username,
> > >   (SELECT COUNT(t.techid)
> > >    FROM ticket t
> > >    WHERE t.techid = u.idnum)
> > > FROM users u;
> >
> > Nope, doesn't work in 6.5.*.  It is there in current sources for the
> > upcoming 7.0 release.  Can't think of any good workaround in 6.5...
> >
> >                         regards, tom lane
> >
> > ************
> 
> i would do an UNION  of the tuples coming from the joint
> with the tuples from "users" who don't have a counterpart in "ticket",
> using "not exists" and adding to them a zero count.
> Hoping it will help
> regards
> P.Jacquot
> 
> 
> ************
> 
> 



В списке pgsql-sql по дате отправления:

Предыдущее
От: wieck@debis.com (Jan Wieck)
Дата:
Сообщение: Re: [SQL] Problem with PL/pgSQL
Следующее
От: "Javi Piñol"
Дата:
Сообщение: POR FIN DINERO REAL SIN TRAMPAS