PG Bug reporting form <noreply@postgresql.org> writes:
> The following script creates a partition table with a foreign key and works
> fine on PostgreSQL 14:
> begin;
> create table a(id int primary key);
> create table b(id int primary key, a_id int not null references a(id))
> partition by range(id);
> create table b_default(like b including defaults including constraints);
> alter table b attach partition b_default default;
> alter table b disable trigger all;
> commit;
> On PostgresSQL 15, it fails on DISABLE TRIGGER ALL:
> ERROR: trigger "RI_ConstraintTrigger_c_20196703" for table "b_default" does
> not exist
It works for me with 15.3. I don't think you are actually running
15.3 (i.e., yesterday's minor release), because this symptom matches
a bug we fixed in 15.3.
regards, tom lane
