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From: vignesh C <vignesh21@gmail.com>
Date: Thu, 13 Feb 2025 11:24:51 +0530
Message-ID: <CALDaNm2Q1rNN-L8u95pjVRw-aBJkcGJh2bHeaPc2bO=mj_1QMA@mail.gmail.com>
Subject: Re: Restrict publishing of partitioned table with a foreign table as partition
To: Shlok Kyal <shlok.kyal.oss@gmail.com>
Cc: =?UTF-8?Q?=C3=81lvaro_Herrera?= <alvherre@alvh.no-ip.org>, 
	Sergey Tatarintsev <s.tatarintsev@postgrespro.ru>, pgsql-hackers@lists.postgresql.org
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On Tue, 11 Feb 2025 at 16:55, Shlok Kyal <shlok.kyal.oss@gmail.com> wrote:
> I have handled the above cases and added tests for the same.

There is a concurrency issue with the patch:
+check_partrel_has_foreign_table(Form_pg_class relform)
+{
+       bool            has_foreign_tbl = false;
+
+       if (relform->relkind == RELKIND_PARTITIONED_TABLE)
+       {
+               List       *relids = NIL;
+
+               relids = find_all_inheritors(relform->oid, NoLock, NULL);
+
+               foreach_oid(relid, relids)
+               {
+                       Relation        rel = table_open(relid,
AccessShareLock);
+
+                       if (RelationGetForm(rel)->relkind ==
RELKIND_FOREIGN_TABLE)
+                               has_foreign_tbl = true;
+
+                       table_close(rel, AccessShareLock);
+
+                       if (has_foreign_tbl)
+                               break;
+               }
+       }
+
+       return has_foreign_tbl;
+}

In an ideal scenario, the creation of a foreign table should fail if
there is an associated publication, as demonstrated below:
CREATE TABLE t(id int) PARTITION BY RANGE(id);
CREATE TABLE part1 PARTITION OF t FOR VALUES FROM (0) TO (5);
CREATE TABLE part2 PARTITION OF t FOR VALUES FROM (5) TO (15)
PARTITION BY RANGE(id);
CREATE PUBLICATION pub1 FOR TABLE t with (publish_via_partition_root = true);

postgres=# CREATE FOREIGN TABLE part22 PARTITION OF part2 FOR VALUES
FROM (10) TO (15) SERVER fdw;
ERROR:  cannot create table foreign partition "part22"
DETAIL:  partition table "part2" is published with option
publish_via_partition_root

Consider a scenario where the publication is being created and after
the check_partrel_has_foreign_table execution is done, concurrently
creation of foreign table is executed, then the creation will be
successful.
postgres=# CREATE FOREIGN TABLE part22 PARTITION OF part2 FOR VALUES
FROM (10) TO (15) SERVER fdw;
CREATE FOREIGN TABLE

I felt the problem here is that you have released the lock:
+                       if (RelationGetForm(rel)->relkind ==
RELKIND_FOREIGN_TABLE)
+                               has_foreign_tbl = true;
+
+                       table_close(rel, AccessShareLock);

We should retain the lock to fix this issue:
+                       if (RelationGetForm(rel)->relkind ==
RELKIND_FOREIGN_TABLE)
+                               has_foreign_tbl = true;
+
+                       table_close(rel, NoLock);

Regards,
Vignesh