Add basic regression tests for semi/antijoin recognition.

Add some simple tests that the planner recognizes all the
standard idioms for SEMI and ANTI joins.  Failure to optimize
in this way won't necessarily cause any visible change in
query results, so check the plans.  We had no similar coverage
before, at least for some variants of antijoin, as noted by
Richard Guo.

Discussion: https://postgr.es/m/CAMbWs4-mvPPCJ1W6iK6dD5HiNwoJdi6mZp=-7mE8N9Sh+cd0tQ@mail.gmail.com
This commit is contained in:
Tom Lane 2022-10-31 19:52:33 -04:00
parent 0717f2fedb
commit 0043aa6b85
2 changed files with 88 additions and 0 deletions

View File

@ -2650,6 +2650,69 @@ WHERE d.f1 IS NULL;
9999
(3 rows)
--
-- basic semijoin and antijoin recognition tests
--
explain (costs off)
select a.* from tenk1 a
where unique1 in (select unique2 from tenk1 b);
QUERY PLAN
------------------------------------------------------------
Hash Semi Join
Hash Cond: (a.unique1 = b.unique2)
-> Seq Scan on tenk1 a
-> Hash
-> Index Only Scan using tenk1_unique2 on tenk1 b
(5 rows)
-- sadly, this is not an antijoin
explain (costs off)
select a.* from tenk1 a
where unique1 not in (select unique2 from tenk1 b);
QUERY PLAN
--------------------------------------------------------
Seq Scan on tenk1 a
Filter: (NOT (hashed SubPlan 1))
SubPlan 1
-> Index Only Scan using tenk1_unique2 on tenk1 b
(4 rows)
explain (costs off)
select a.* from tenk1 a
where exists (select 1 from tenk1 b where a.unique1 = b.unique2);
QUERY PLAN
------------------------------------------------------------
Hash Semi Join
Hash Cond: (a.unique1 = b.unique2)
-> Seq Scan on tenk1 a
-> Hash
-> Index Only Scan using tenk1_unique2 on tenk1 b
(5 rows)
explain (costs off)
select a.* from tenk1 a
where not exists (select 1 from tenk1 b where a.unique1 = b.unique2);
QUERY PLAN
------------------------------------------------------------
Hash Anti Join
Hash Cond: (a.unique1 = b.unique2)
-> Seq Scan on tenk1 a
-> Hash
-> Index Only Scan using tenk1_unique2 on tenk1 b
(5 rows)
explain (costs off)
select a.* from tenk1 a left join tenk1 b on a.unique1 = b.unique2
where b.unique2 is null;
QUERY PLAN
------------------------------------------------------------
Hash Anti Join
Hash Cond: (a.unique1 = b.unique2)
-> Seq Scan on tenk1 a
-> Hash
-> Index Only Scan using tenk1_unique2 on tenk1 b
(5 rows)
--
-- regression test for proper handling of outer joins within antijoins
--

View File

@ -624,6 +624,31 @@ LEFT JOIN (
) AS d ON (a.f1 = d.f1)
WHERE d.f1 IS NULL;
--
-- basic semijoin and antijoin recognition tests
--
explain (costs off)
select a.* from tenk1 a
where unique1 in (select unique2 from tenk1 b);
-- sadly, this is not an antijoin
explain (costs off)
select a.* from tenk1 a
where unique1 not in (select unique2 from tenk1 b);
explain (costs off)
select a.* from tenk1 a
where exists (select 1 from tenk1 b where a.unique1 = b.unique2);
explain (costs off)
select a.* from tenk1 a
where not exists (select 1 from tenk1 b where a.unique1 = b.unique2);
explain (costs off)
select a.* from tenk1 a left join tenk1 b on a.unique1 = b.unique2
where b.unique2 is null;
--
-- regression test for proper handling of outer joins within antijoins
--