Re: BUG #15724: Can't create foreign table as partition - Mailing list pgsql-bugs

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/
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