Horton

Re: [PATCH] GROUP BY ALL

Peter Eisentraut <peter@eisentraut.org>

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From: Peter Eisentraut <peter@eisentraut.org>
To: Tom Lane <tgl@sss.pgh.pa.us>
Cc: pgsql-hackers <pgsql-hackers@postgresql.org>, "David G. Johnston" <david.g.johnston@gmail.com>, David Christensen <david@pgguru.net>, Jelte Fennema-Nio <postgres@jeltef.nl>
Date: 2025-09-26T15:54:18Z
Lists: pgsql-hackers

Commits

Same data as JSON: GET /api/v1/messages/:b64id/commits the thread's linked commits as JSON, with link sources. API reference →

  1. Add GROUP BY ALL.

  2. Refactor to avoid code duplication in transformPLAssignStmt.

  3. Fix missed copying of groupDistinct in transformPLAssignStmt.

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