Re: BUG #15724: Can't create foreign table as partition

From: Alvaro Herrera <alvherre(at)2ndquadrant(dot)com>
To: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Cc: stepya(at)ukr(dot)net, pgsql-bugs(at)lists(dot)postgresql(dot)org
Subject: Re: BUG #15724: Can't create foreign table as partition
Date: 2019-03-29 14:10:50
Message-ID: 20190329141050.GA22336@alvherre.pgsql
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On 2019-Mar-29, Tom Lane wrote:

> Alvaro Herrera <alvherre(at)2ndquadrant(dot)com> writes:
> > On 2019-Mar-29, PG Bug reporting form wrote:
> >> SQL Error [42809]: ERROR: cannot create index on foreign table "msg_json"
>
> > Ah, this is because we try to propagate the index to the partition,
> > which appears to be less than desirable. Maybe this patch fixes it? I
> > have not tested it, only verified that it compiles.
>
> Really, what *ought* to happen in such a case is that the FDW gets
> told about the command and has the opportunity to try to do something
> corresponding to making an index. But that smells like a new feature
> rather than a bug fix.

ENOTIME to get it done for pg12 I suppose (I don't know a thing about
postgres_fdw or the FDW API layer).

> I'm not sure that just forcing the case to be a no-op is wise.
> What if the index is supposed to be UNIQUE?

Hmm, good point, that should be disallowed too.

--
Álvaro Herrera https://www.2ndQuadrant.com/
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services

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