Re: Restrict publishing of partitioned table with a foreign table as partition

On Tue, May 20, 2025 at 2:33 AM Shlok Kyal <shlok(dot)kyal(dot)oss(at)gmail(dot)com> wrote:
>
> This approach seems better to me. I have created a patch with the
> above approach.
>
> Thanks and Regards,
> Shlok Kyal

Some quick comments on the patch:
1. In doc/src/sgml/ref/create_subscription.sgml:
+ has partitioned table with foreign table as partition. If this scenario is
+ detected we ERROR is logged to the user.
+ </para>
+

Should be: "If this scenario is detected an ERROR is logged to the
user." (remove "we").

In src/backend/commands/subscriptioncmds.c:
2. The comment header:
+ * This function is in charge of detecting if publisher with
+ * publish_via_partition_root=true publishes a partitioned table that has a
+ * foreign table as a partition.

Add "and throw an error if found" at the end of that sentence to
correctly describe what the function does.

3.
+ appendStringInfoString(&cmd,
+ "SELECT DISTINCT P.pubname AS pubname "
+ "from pg_catalog.pg_publication p, LATERAL "
+ "pg_get_publication_tables(p.pubname) gpt, LATERAL "
+ "pg_partition_tree(gpt.relid) gt JOIN
pg_catalog.pg_foreign_table ft ON "
+ "ft.ftrelid = gt.relid WHERE p.pubviaroot
= true AND p.pubname IN (");

use FROM rather than from to maintain SQL style consistency.

4.
+ errdetail_plural("The subscription being created on a
publication (%s) with publish_via_root_partition = true and contains
partitioned tables with foreign table as partition ",
+ "The subscription being created on
publications (%s) with publish_via_root_partition = true and contains
partitioned tables with foreign table as partition ",
+ list_length(publist), pubnames->data),

I think you meant "publish_via_partition_root" here and not
"publish_via_root_partition ".

regards,
Ajin Cherian
Fujitsu Australia