Re: [PATCH] GROUP BY ALL

On Fri, Sep 26, 2025 at 10:54 AM Peter Eisentraut <peter@eisentraut.org> wrote:
>
> On 26.09.25 16:11, Tom Lane wrote:
> > Peter Eisentraut <peter@eisentraut.org> writes:
> >> The initially proposed patch appears to have the right idea overall.
> >> But it does not handle more complex cases like
> >>       SELECT a, SUM(b)+a FROM t1 GROUP BY ALL;
> >
> >> (For explanation:  GROUP BY ALL expands to all select list entries that
> >> do not contain aggregates.  So the above would expand to
> >>       SELECT a, SUM(b)+a FROM t1 GROUP BY a;
> >> which should then be rejected based on the existing rules.)
> >
> > I thought I understood this definition, up till your last
> > comment.  What's invalid about that expanded query?
> >
> > regression=# create table t1 (a int, b int);
> > CREATE TABLE
> > regression=# SELECT a, SUM(b)+a FROM t1 GROUP BY a;
> >   a | ?column?
> > ---+----------
> > (0 rows)
>
> This was a sloppy example.  Here is a better one:
>
>      create table t1 (a int, b int, c int);
>
>      select a, sum(b)+c from t1 group by all;
>
> This is equivalent to
>
>      select a, sum(b)+c from t1 group by a;
>
> which would be rejected as
>
>      ERROR:  column "t1.c" must appear in the GROUP BY clause or be used
>      in an aggregate function

Verified that with v2 that this is what happens in this case; will
include this and whatever other feedback there is in a v3.