| From: | Alvaro Herrera <alvherre(at)2ndquadrant(dot)com> |
|---|---|
| To: | Amit Langote <amitlangote09(at)gmail(dot)com> |
| Cc: | Pavan Deolasee <pavan(dot)deolasee(at)gmail(dot)com>, stepya(at)ukr(dot)net, PostgreSQL mailing lists <pgsql-bugs(at)lists(dot)postgresql(dot)org> |
| Subject: | Re: BUG #15724: Can't create foreign table as partition |
| Date: | 2019-06-26 20:18:43 |
| Message-ID: | 20190626201843.GA18542@alvherre.pgsql |
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| Lists: | pgsql-bugs |
On 2019-Jun-26, Amit Langote wrote:
> On Tue, Jun 25, 2019 at 9:57 PM Alvaro Herrera <alvherre(at)2ndquadrant(dot)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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