Avoid using EXPLAIN VERBOSE when temp tables are involved in a test case.

In XL, the temporary schema may change in different regression runs. Hence we
must not print the schema in expected output. Change the sql as well as
expected output file for the 'join' test case.

diff --git a/src/test/regress/expected/join.out b/src/test/regress/expected/join.out
index 1762b3cbf07dc7643f58571f544d2b4035077b21..5c579e9c3abca356f8ac520c07eded4a61d0b5af 100644 (file)
--- a/src/test/regress/expected/join.out
+++ b/src/test/regress/expected/join.out
@@ -2979,18 +2979,28 @@ explain (costs off)
 select * from
   tenk1, int8_tbl a, int8_tbl b
 where thousand = a.q1 and tenthous = b.q1 and a.q2 = 1 and b.q2 = 2;
-                                QUERY PLAN                                 
----------------------------------------------------------------------------
- Remote Subquery Scan on all (datanode_1,datanode_2)
-   ->  Nested Loop
-         ->  Seq Scan on int8_tbl b
-               Filter: (q2 = 2)
-         ->  Nested Loop
+                       QUERY PLAN                        
+---------------------------------------------------------
+ Remote Fast Query Execution
+   Node/s: datanode_1, datanode_2
+   ->  Merge Join
+         Merge Cond: (tenk1.thousand = a.q1)
+         ->  Sort
+               Sort Key: tenk1.thousand
+               ->  Merge Join
+                     Merge Cond: (tenk1.tenthous = b.q1)
+                     ->  Sort
+                           Sort Key: tenk1.tenthous
+                           ->  Seq Scan on tenk1
+                     ->  Sort
+                           Sort Key: b.q1
+                           ->  Seq Scan on int8_tbl b
+                                 Filter: (q2 = 2)
+         ->  Sort
+               Sort Key: a.q1
                ->  Seq Scan on int8_tbl a
                      Filter: (q2 = 1)
-               ->  Index Scan using tenk1_thous_tenthous on tenk1
-                     Index Cond: ((thousand = a.q1) AND (tenthous = b.q1))
-(9 rows)
+(19 rows)
 
 --
 -- test a corner case in which we shouldn't apply the star-schema optimization
@@ -4027,10 +4037,12 @@ explain (verbose, costs off)
   select a.q2, b.q1
     from int8_tbl a left join int8_tbl b on a.q2 = coalesce(b.q1, 1)
     where coalesce(b.q1, 1) > 0;
-                          QUERY PLAN                           
----------------------------------------------------------------
- Remote Subquery Scan on all (datanode_1)
+                                                                         QUERY PLAN                                                                         
+------------------------------------------------------------------------------------------------------------------------------------------------------------
+ Remote Fast Query Execution
    Output: a.q2, b.q1
+   Node/s: datanode_1
+   Remote query: SELECT a.q2, b.q1 FROM (int8_tbl a LEFT JOIN int8_tbl b ON ((a.q2 = COALESCE(b.q1, (1)::bigint)))) WHERE (COALESCE(b.q1, (1)::bigint) > 0)
    ->  Merge Left Join
          Output: a.q2, b.q1
          Merge Cond: (a.q2 = (COALESCE(b.q1, '1'::bigint)))
@@ -4045,7 +4057,7 @@ explain (verbose, costs off)
                Sort Key: (COALESCE(b.q1, '1'::bigint))
                ->  Seq Scan on public.int8_tbl b
                      Output: b.q1, COALESCE(b.q1, '1'::bigint)
-(16 rows)
+(18 rows)
 
 select a.q2, b.q1
   from int8_tbl a left join int8_tbl b on a.q2 = coalesce(b.q1, 1)
@@ -4079,54 +4091,43 @@ INSERT INTO b VALUES (0, 0), (1, NULL);
 INSERT INTO c VALUES (0), (1);
 INSERT INTO d VALUES (1,3), (2,2), (3,1);
 -- all three cases should be optimizable into a simple seqscan
-explain (verbose true, costs false, nodes false) SELECT a.* FROM a LEFT JOIN b ON a.b_id = b.id;
-          QUERY PLAN           
--------------------------------
+explain (verbose false, costs false, nodes false) SELECT a.* FROM a LEFT JOIN b ON a.b_id = b.id;
+         QUERY PLAN          
+-----------------------------
  Remote Subquery Scan on all
-   Output: a.id, a.b_id
-   ->  Seq Scan on pg_temp_5.a
-         Output: a.id, a.b_id
-(4 rows)
+   ->  Seq Scan on a
+(2 rows)
 
-explain (verbose true, costs false, nodes false) SELECT b.* FROM b LEFT JOIN c ON b.c_id = c.id;
-          QUERY PLAN           
--------------------------------
+explain (verbose false, costs false, nodes false) SELECT b.* FROM b LEFT JOIN c ON b.c_id = c.id;
+         QUERY PLAN          
+-----------------------------
  Remote Subquery Scan on all
-   Output: b.id, b.c_id
-   ->  Seq Scan on pg_temp_5.b
-         Output: b.id, b.c_id
-(4 rows)
+   ->  Seq Scan on b
+(2 rows)
 
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
   SELECT a.* FROM a LEFT JOIN (b left join c on b.c_id = c.id)
   ON (a.b_id = b.id);
-          QUERY PLAN           
--------------------------------
+         QUERY PLAN          
+-----------------------------
  Remote Subquery Scan on all
-   Output: a.id, a.b_id
-   ->  Seq Scan on pg_temp_5.a
-         Output: a.id, a.b_id
-(4 rows)
+   ->  Seq Scan on a
+(2 rows)
 
 -- check optimization of outer join within another special join
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
 select id from a where id in (
        select b.id from b left join c on b.id = c.id
 );
-                QUERY PLAN                 
--------------------------------------------
+            QUERY PLAN            
+----------------------------------
  Remote Subquery Scan on all
-   Output: a.id
    ->  Hash Semi Join
-         Output: a.id
          Hash Cond: (a.id = b.id)
-         ->  Seq Scan on pg_temp_5.a
-               Output: a.id, a.b_id
+         ->  Seq Scan on a
          ->  Hash
-               Output: b.id
-               ->  Seq Scan on pg_temp_5.b
-                     Output: b.id
-(11 rows)
+               ->  Seq Scan on b
+(6 rows)
 
 -- check that join removal works for a left join when joining a subquery
 -- that is guaranteed to be unique by its GROUP BY clause
@@ -4221,18 +4222,15 @@ select p.* from parent p left join child c on (p.k = c.k) order by 1,2;
  3 | 30
 (3 rows)
 
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
   select p.* from parent p left join child c on (p.k = c.k) order by 1,2;
-                 QUERY PLAN                 
---------------------------------------------
+            QUERY PLAN            
+----------------------------------
  Remote Subquery Scan on all
-   Output: p.k, p.pd
    ->  Sort
-         Output: p.k, p.pd
          Sort Key: p.k, p.pd
-         ->  Seq Scan on pg_temp_5.parent p
-               Output: p.k, p.pd
-(7 rows)
+         ->  Seq Scan on parent p
+(4 rows)
 
 -- this case is not
 select p.*, linked from parent p
@@ -4245,22 +4243,18 @@ select p.*, linked from parent p
  3 | 30 | 
 (3 rows)
 
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
   select p.*, linked from parent p
     left join (select c.*, true as linked from child c) as ss
     on (p.k = ss.k) order by p.k;
-                             QUERY PLAN                             
---------------------------------------------------------------------
+                        QUERY PLAN                        
+----------------------------------------------------------
  Remote Subquery Scan on all
-   Output: p.k, p.pd, true
    ->  Merge Left Join
-         Output: p.k, p.pd, (true)
          Merge Cond: (p.k = c.k)
-         ->  Index Scan using parent_pkey on pg_temp_5.parent p
-               Output: p.k, p.pd
-         ->  Index Only Scan using child_k_key on pg_temp_5.child c
-               Output: c.k, true
-(9 rows)
+         ->  Index Scan using parent_pkey on parent p
+         ->  Index Only Scan using child_k_key on child c
+(5 rows)
 
 -- check for a 9.0rc1 bug: join removal breaks pseudoconstant qual handling
 select p.* from
@@ -4270,22 +4264,18 @@ select p.* from
 ---+----
 (0 rows)
 
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
 select p.* from
   parent p left join child c on (p.k = c.k)
   where p.k = 1 and p.k = 2;
-                                                     QUERY PLAN                                                      
----------------------------------------------------------------------------------------------------------------------
+                      QUERY PLAN                      
+------------------------------------------------------
  Remote Fast Query Execution
-   Output: p.k, p.pd
-   Remote query: SELECT p.k, p.pd FROM (parent p LEFT JOIN child c ON ((p.k = c.k))) WHERE ((p.k = 1) AND (p.k = 2))
    ->  Result
-         Output: p.k, p.pd
          One-Time Filter: false
-         ->  Index Scan using parent_pkey on pg_temp_12.parent p
-               Output: p.k, p.pd
-               Index Cond: (p.k = 1)
-(9 rows)
+         ->  Index Scan using parent_pkey on parent p
+               Index Cond: (k = 1)
+(5 rows)
 
 select p.* from
   (parent p left join child c on (p.k = c.k)) join parent x on p.k = x.k
@@ -4294,19 +4284,16 @@ select p.* from
 ---+----
 (0 rows)
 
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
 select p.* from
   (parent p left join child c on (p.k = c.k)) join parent x on p.k = x.k
   where p.k = 1 and p.k = 2;
-                                                                      QUERY PLAN                                                                      
-------------------------------------------------------------------------------------------------------------------------------------------------------
+           QUERY PLAN           
+--------------------------------
  Remote Fast Query Execution
-   Output: p.k, p.pd
-   Remote query: SELECT p.k, p.pd FROM ((parent p LEFT JOIN child c ON ((p.k = c.k))) JOIN parent x ON ((p.k = x.k))) WHERE ((p.k = 1) AND (p.k = 2))
    ->  Result
-         Output: p.k, p.pd
          One-Time Filter: false
-(6 rows)
+(3 rows)
 
 -- bug 5255: this is not optimizable by join removal
 begin;

diff --git a/src/test/regress/sql/join.sql b/src/test/regress/sql/join.sql
index 6a3a63981d85cd751e4830b60c27f061aaaa6a78..09f14a0070cd42d526b3ad125b21058bc7de1d43 100644 (file)
--- a/src/test/regress/sql/join.sql
+++ b/src/test/regress/sql/join.sql
@@ -1279,14 +1279,14 @@ INSERT INTO c VALUES (0), (1);
 INSERT INTO d VALUES (1,3), (2,2), (3,1);
 
 -- all three cases should be optimizable into a simple seqscan
-explain (verbose true, costs false, nodes false) SELECT a.* FROM a LEFT JOIN b ON a.b_id = b.id;
-explain (verbose true, costs false, nodes false) SELECT b.* FROM b LEFT JOIN c ON b.c_id = c.id;
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false) SELECT a.* FROM a LEFT JOIN b ON a.b_id = b.id;
+explain (verbose false, costs false, nodes false) SELECT b.* FROM b LEFT JOIN c ON b.c_id = c.id;
+explain (verbose false, costs false, nodes false)
   SELECT a.* FROM a LEFT JOIN (b left join c on b.c_id = c.id)
   ON (a.b_id = b.id);
 
 -- check optimization of outer join within another special join
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
 select id from a where id in (
        select b.id from b left join c on b.id = c.id
 );
@@ -1335,14 +1335,14 @@ insert into child values (1, 100), (4, 400);
 
 -- this case is optimizable
 select p.* from parent p left join child c on (p.k = c.k) order by 1,2;
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
   select p.* from parent p left join child c on (p.k = c.k) order by 1,2;
 
 -- this case is not
 select p.*, linked from parent p
   left join (select c.*, true as linked from child c) as ss
   on (p.k = ss.k) order by p.k;
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
   select p.*, linked from parent p
     left join (select c.*, true as linked from child c) as ss
     on (p.k = ss.k) order by p.k;
@@ -1351,7 +1351,7 @@ explain (verbose true, costs false, nodes false)
 select p.* from
   parent p left join child c on (p.k = c.k)
   where p.k = 1 and p.k = 2;
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
 select p.* from
   parent p left join child c on (p.k = c.k)
   where p.k = 1 and p.k = 2;
@@ -1359,7 +1359,7 @@ select p.* from
 select p.* from
   (parent p left join child c on (p.k = c.k)) join parent x on p.k = x.k
   where p.k = 1 and p.k = 2;
-explain (verbose true, costs false, nodes false)
+explain (verbose false, costs false, nodes false)
 select p.* from
   (parent p left join child c on (p.k = c.k)) join parent x on p.k = x.k
   where p.k = 1 and p.k = 2;