From: Alvaro Herrera <alvherre@2ndquadrant.com>
To: Amit Langote <amitlangote09@gmail.com>
Cc: Pavan Deolasee <pavan.deolasee@gmail.com>
Cc: stepya@ukr.net, PostgreSQL mailing lists <pgsql-bugs@lists.postgresql.org>
Subject: Re: BUG #15724: Can't create foreign table as partition
Date: Wed, 26 Jun 2019 16:18:43 -0400
Message-ID: <20190626201843.GA18542@alvherre.pgsql> (raw)
In-Reply-To: <CA+HiwqEoxH0DFhVeZ-no4iy1kEb41Dus=Yj-8oVAx=sSfTSJNA@mail.gmail.com>
On 2019-Jun-26, Amit Langote wrote:
> On Tue, Jun 25, 2019 at 9:57 PM Alvaro Herrera <alvherre@2ndquadrant.com> wrote:
> > Because if you do that, you might build a few indexes on regular
> > partitions before coming across a foreign one, which is very unfriendly.
> > I'll add a comment to this effect.
>
> Hmm, why would other partitions be involved?
> AttachPartitionEnsureIndexes() only considers the partition being
> attached, like DefineRelation only considers the partition being
> created.
Sorry, I meant other *indexes*. You might build a few regular
(non-unique) indexes before coming across a unique index, at which point
you throw the error.
--
Álvaro Herrera https://www.2ndQuadrant.com/
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services
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Cc: alvherre@2ndquadrant.com, amitlangote09@gmail.com, pavan.deolasee@gmail.com, pgsql-bugs@lists.postgresql.org
Subject: Re: BUG #15724: Can't create foreign table as partition
In-Reply-To: <20190626201843.GA18542@alvherre.pgsql>
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