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
