From: Tom Lane <tgl@sss.pgh.pa.us>
To: "David G. Johnston" <david.g.johnston@gmail.com>
cc: "luy70@psu.edu" <luy70@psu.edu>,
        "pgsql-bugs@lists.postgresql.org" <pgsql-bugs@lists.postgresql.org>
Subject: Re: BUG #17150: Unexpected outputs from the query
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 <CAKFQuwaxSxw+T3dmyOoNLW=zrL5yPz3_ecaVMdVyihYCTMa9rw@mail.gmail.com>
References: <17150-d22e14860d93d32e@postgresql.org>
 <CAKFQuwaxSxw+T3dmyOoNLW=zrL5yPz3_ecaVMdVyihYCTMa9rw@mail.gmail.com>
Comments: In-reply-to "David G. Johnston" <david.g.johnston@gmail.com>
	message dated "Tue, 17 Aug 2021 11:27:10 -0700"
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Date: Tue, 17 Aug 2021 14:56:43 -0400
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"David G. Johnston" <david.g.johnston@gmail.com> writes:
> On Tuesday, August 17, 2021, PG Bug reporting form <noreply@postgresql.org>
> wrote:
>> This
>> unexpected return can be fixed by removing "ORDER BY ( SELECT COUNT ( v1 )
>> )", then the query returns sum="0" as expected.

> Well, PostgreSQL cannot remove the order by otherwise it would be a
> different query.  So your suggestion is spot on, and the user should
> probably do that, but it doesn’t seem like a bug.

Yeah.  PG interprets

	SELECT x FROM v2 ORDER BY (SELECT COUNT(v1))

to behave the same as

	SELECT x, (SELECT COUNT(v1)) FROM v2 ORDER BY 2

(modulo the fact that the ORDER BY column won't be output),
and then it turns out that that's effectively the same as

	SELECT x, COUNT(v1) FROM v2 ORDER BY 2

the reason being that since v1 is a variable of the outer query,
the aggregate is considered to be an aggregate of the outer query
*not* the sub-select.  (That's required by the SQL standard.)
So at this point you have an aggregated query that is certain
to return 1 row, not more or less, regardless of how many rows
are returned by v2.

			regards, tom lane