Attempt to stabilize new window agg regression test

This test was recently added in 3900a02c9.  It appears to be unstable in
regards to the join order presumably due to the relations at either side
of the join being equal in side.  Here we add a qual to make one of them
smaller so the planner is more likely to choose to hash the smaller of the
two.

Reported-by: Nathan Bossart, Tom Lane
Discussion: https://www.postgr.es/m/20230803235403.GC1238296@nathanxps13

diff --git a/src/test/regress/expected/window.out b/src/test/regress/expected/window.out
index 2628033327d8f82397f930d6bee3df24cd48640d..69a38df10b6ff4746b4f9bfe30b17481b1de177e 100644 (file)
--- a/src/test/regress/expected/window.out
+++ b/src/test/regress/expected/window.out
@@ -4834,17 +4834,19 @@ LIMIT 1;
 EXPLAIN (COSTS OFF)
 SELECT COUNT(*) OVER ()
 FROM tenk1 t1 INNER JOIN tenk1 t2 ON t1.unique1 = t2.tenthous
+WHERE t2.two = 1
 LIMIT 1;
-                                   QUERY PLAN                                   
---------------------------------------------------------------------------------
+                            QUERY PLAN                             
+-------------------------------------------------------------------
  Limit
    ->  WindowAgg
          ->  Hash Join
                Hash Cond: (t1.unique1 = t2.tenthous)
                ->  Index Only Scan using tenk1_unique1 on tenk1 t1
                ->  Hash
-                     ->  Index Only Scan using tenk1_thous_tenthous on tenk1 t2
-(7 rows)
+                     ->  Seq Scan on tenk1 t2
+                           Filter: (two = 1)
+(8 rows)
 
 -- Ensure we get a cheap total plan.  This time use UNBOUNDED FOLLOWING, which
 -- needs to read all join rows to output the first WindowAgg row.

diff --git a/src/test/regress/sql/window.sql b/src/test/regress/sql/window.sql
index 4789de09376b3844a8e8b4e838b49d0ee93dd98f..9113a92ae0b1e1ac66e6b795f3324827dc6371bc 100644 (file)
--- a/src/test/regress/sql/window.sql
+++ b/src/test/regress/sql/window.sql
@@ -1734,6 +1734,7 @@ LIMIT 1;
 EXPLAIN (COSTS OFF)
 SELECT COUNT(*) OVER ()
 FROM tenk1 t1 INNER JOIN tenk1 t2 ON t1.unique1 = t2.tenthous
+WHERE t2.two = 1
 LIMIT 1;
 
 -- Ensure we get a cheap total plan.  This time use UNBOUNDED FOLLOWING, which