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From: David Christensen <david@pgguru.net>
Date: Fri, 26 Sep 2025 10:59:26 -0500
Message-ID: 
 <CAHM0NXjic04De6eMcYLZ+-u1t-7uBEU6xXgatz6eb-MECytetg@mail.gmail.com>
Subject: Re: [PATCH] GROUP BY ALL
To: Peter Eisentraut <peter@eisentraut.org>
Cc: Tom Lane <tgl@sss.pgh.pa.us>,
 pgsql-hackers <pgsql-hackers@postgresql.org>,
	"David G. Johnston" <david.g.johnston@gmail.com>,
 Jelte Fennema-Nio <postgres@jeltef.nl>
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On Fri, Sep 26, 2025 at 10:54=E2=80=AFAM 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 tha=
t
> >> 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=3D# create table t1 (a int, b int);
> > CREATE TABLE
> > regression=3D# 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.