Re[2]: BUG #15724: Can't create foreign table as partition

From: Stepan Yankevych <stepya(at)ukr(dot)net>
To: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Cc: Alvaro Herrera <alvherre(at)2ndquadrant(dot)com>, pgsql-bugs(at)lists(dot)postgresql(dot)org
Subject: Re[2]: BUG #15724: Can't create foreign table as partition
Date: 2019-03-29 16:25:42
Message-ID: 1553876622.668408412.24u6zti5@frv53.fwdcdn.com
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>> 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?Anyway it would be better to have no-op in 11.3 than having it not working. We just can put comment in the doc like. Foreign Server should take care about indexing foreign table as partition.

--- Original message ---
From: "Tom Lane" <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Date: 29 March 2019, 16:01:41

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.

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?

regards, tom lane

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