Re: pg_get_publication_tables() output duplicate relid

On Tue, Dec 7, 2021 at 3:20 PM Amit Kapila <amit(dot)kapila16(at)gmail(dot)com> wrote:
> On Tue, Dec 7, 2021 at 10:52 AM Amit Langote <amitlangote09(at)gmail(dot)com> wrote:
> > So if a partition is
> > explicitly present in a publication with a filter defined on it, I
> > suppose following are the possible scenarios:
> >
> > * publish_via_partition_root=true
> >
> > 1. If the root table is not present in the publication (perhaps
> > unlikely), use the partition's filter because there's nothing else to
> > consider.
> >
> > 2. If the root table is present in the publication (very likely), use
> > the root's filter, ignoring/overriding the partition's own if any.
> >
> > * publish_via_partition_root=false
> >
> > 1. If the root table is not present in the publication, use the
> > partition's filter because there's nothing else to consider.
> >
> > 2. If the root table is present in the publication, use the
> > partition's own filter if present, else root's.
>
> I have not tested/checked each of these scenarios individually but I
> think it is something like, if publish_via_partition_root is false
> then we will always try to use partitions row filter and if it is not
> there then we don't use any filter. Similarly, if
> publish_via_partition_root is true, then we always try to use a
> partitioned table row filter and if it is not there (either because
> the partitioned table is not part of the publication or because there
> is no filter defined for it) then we don't use any filter.

Okay, thanks for the explanation.

--
Amit Langote
EDB: http://www.enterprisedb.com