Re: BUG #15724: Can't create foreign table as partition
| От | Pavan Deolasee |
|---|---|
| Тема | Re: BUG #15724: Can't create foreign table as partition |
| Дата | |
| Msg-id | CABOikdP5KPfJOpWVtU-XvHoXbHYntJ47+sCf_r6Z5WRCZsArFg@mail.gmail.com обсуждение исходный текст |
| Ответ на | Re: BUG #15724: Can't create foreign table as partition (Alvaro Herrera <alvherre@2ndquadrant.com>) |
| Ответы |
Re: BUG #15724: Can't create foreign table as partition
|
| Список | pgsql-bugs |
On Thu, Jun 20, 2019 at 6:07 AM Alvaro Herrera <alvherre@2ndquadrant.com> wrote:
With this patch, an index creation will no longer fail in the presence
of a partition that is a foreign table, as long as the index is not a
constraint index (not unique, not primary key). Conversely,
creating/attaching a partition that is a foreign table does not fail if
the partitioned table only has non-constraint indexes.
Like others suggested above, I also think that we should make this is a no-op on the foreign tables i.e. not fail even when there exists a UNIQUE or PRIMARY KEY on the parent table. We simply assume that the appropriate constraints will be defined on the foreign side and violations will be caught. This is same as CHECK constraints on the foreign partitions, that we assume the foreign server will enforce.
Thanks,
Pavan
Pavan Deolasee http://www.2ndQuadrant.com/
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