Re: 7.3 "group by" issue

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От Chad Thompson
Тема Re: 7.3 "group by" issue
Дата
Msg-id 04f401c2d9e3$f24c1070$32021aac@chad
обсуждение исходный текст
Ответ на Re: 7.3 "group by" issue  ("Dan Langille" <dan@langille.org>)
Ответы Re: 7.3 "group by" issue  ("Dan Langille" <dan@langille.org>)
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> On 21 Feb 2003 at 19:18, Gaetano Mendola wrote:
>
> > > Hi folks,
> > >
> > > This query:
> > >
> > > SELECT element_id as wle_element_id, COUNT(watch_list_id)
> > >     FROM watch_list JOIN watch_list_element
> > >         ON watch_list.id      = watch_list_element.watch_list_id
> > >        AND watch_list.user_id = 1
> > >   GROUP BY watch_list_element.element_id
> >
> > Try:
> >
> > SELECT element_id as wle_element_id, COUNT(watch_list_id)
> >      FROM watch_list JOIN watch_list_element
> >          ON watch_list.id      = watch_list_element.watch_list_id
> > WHERE
> >     watch_list.user_id = 1
> >    GROUP BY watch_list_element.element_id
>
> ERROR:  Attribute unnamed_join.element_id must be GROUPed or used in
> an aggregate function
>

I think that the wrong problem was solved here.  Items in the order by
clause must be in the target list.

heres what it says in the docs
*The ORDER BY clause specifies the sort order:

*SELECT select_list
*     FROM table_expression
*     ORDER BY column1 [ASC | DESC] [, column2 [ASC | DESC] ...]
*column1, etc., refer to select list columns. These can be either the output
name of a column (see Section 4.3.2) or the number of a column. Some
examples:

Note that "column1, etc., refer to select list"

HTH

Chad




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