Re: BUG #18652: Planner can not find pathkey item to sort for query with expression and expression index

From: Tender Wang <tndrwang(at)gmail(dot)com>
To: Andrei Lepikhov <lepihov(at)gmail(dot)com>
Cc: Richard Guo <guofenglinux(at)gmail(dot)com>, exclusion(at)gmail(dot)com, pgsql-bugs(at)lists(dot)postgresql(dot)org, Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Subject: Re: BUG #18652: Planner can not find pathkey item to sort for query with expression and expression index
Date: 2024-10-10 09:02:16
Message-ID: CAHewXNmFmXFxK7bsgAhqv_ToGxK4y7niNVL3GAzgWEOKuiAg5w@mail.gmail.com
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Andrei Lepikhov <lepihov(at)gmail(dot)com> 于2024年10月10日周四 14:50写道:

> On 10/10/24 10:52, Richard Guo wrote:
> > On Thu, Oct 10, 2024 at 5:43 AM Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us> wrote:
> >> It looks like we are generating a Path tree in which one of the
> >> inputs to a MergeAppend is a plain unsorted seqscan, which'd
> >> be all right except it doesn't expose the required sort value
> >> in its targetlist.
> >
> > Correct. In addition, find_computable_ec_member() fails to find a
> > computable expression from its targetlist.
> > For the indexscan path, find_computable_ec_member() is able to find
> > (t.i + 0) which can be computed from its tlist item 't.i'.
> >
> > For the seqscan path, though, find_computable_ec_member() is not able
> > to find ((t_1.i + 1) + 0) from its tlist item '(t_1.i + 1)'.
> >
> > I think this is because find_computable_ec_member() only tries to
> > match Vars. Maybe we should teach it to also match OpExprs?
> Looking into that case, I don't understand only one thing:
> generate_orderedappend_paths decided to try MergeAppend; the
> create_append_path routine added the Sort cost, but the Sort node
> itself wasn't added. Maybe the origin problem is the lack of feasibility
> examinations?
>
>
Yeah, I'm also curious why only cost Sort but not adding Sort node in
create_merge_append_path().
The comments say that " We'll need to insert a Sort node, so include cost
for that". Does another place
insert the Sort node?

--
Thanks,
Tender Wang

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