Re: BUG #18953: Planner fails to build plan for complex query with LATERAL references
От | Tom Lane |
---|---|
Тема | Re: BUG #18953: Planner fails to build plan for complex query with LATERAL references |
Дата | |
Msg-id | 835659.1750177751@sss.pgh.pa.us обсуждение исходный текст |
Ответ на | BUG #18953: Planner fails to build plan for complex query with LATERAL references (PG Bug reporting form <noreply@postgresql.org>) |
Список | pgsql-bugs |
Richard Guo <guofenglinux@gmail.com> writes: > On Tue, Jun 17, 2025 at 12:24 AM Tom Lane <tgl@sss.pgh.pa.us> wrote: >> In the only test case I have that exercises this >> logic, both paths are marked with the required_outer anyway, so it >> doesn't seem to matter. If we can find a case where that's not so, >> then probably we should do it the other way. > I tried to find such a case but didn't succeed; though I suspect > that's simply because I haven't tried hard enough. Conceptually, I'm > thinking of a query where 1) A laterally references both B and C while > B does not reference any other relation, and 2) A has a PHV parameter > whose ph_eval_at includes both B and C. In such a query, the B/A join > path is parameterized by C, while its left path B is not parameterized > at all. In order to add the PHV to B's tlist, we'll need to consider > the join path's required-outer rels not just those of the left path. After thinking about this for awhile, I'm not seeing how that could happen. A PHV with ph_eval_at exceeding its syntactic scope could only be created via something like B left join lateral (select coalesce(B.x, ...) as Q from D) C that is we need a non-strict targetlist expression that references something outside the sub-select proper. In this case all paths created for C will have required_outer mentioning B, because there's no way to compute C's reltarget without an outer reference to B. This might be embedded in ... join A on A.y = C.Q which gives rise to the situation you describe --- but there's no possibility of the planner trying to do this in the order C join (B join A) because it will think that all paths for C require B on the outside. It will only consider B join (C join A) and the path for C will show B as required_outer. So I'm inclined to leave that code as I had it. It's notationally a bit simpler and it doesn't require assuming that we can ignore the path's required_outer marking at this stage. If I'm wrong, someone will eventually find a counterexample and we can fix it then; the changes won't be large. regards, tom lane
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