Teach reparameterize_path() to handle AppendPaths.

If we're inside a lateral subquery, there may be no unparameterized paths
for a particular child relation of an appendrel, in which case we *must*
be able to create similarly-parameterized paths for each other child
relation, else the planner will fail with "could not devise a query plan
for the given query".  This means that there are situations where we'd
better be able to reparameterize at least one path for each child.

This calls into question the assumption in reparameterize_path() that
it can just punt if it feels like it.  However, the only case that is
known broken right now is where the child is itself an appendrel so that
all its paths are AppendPaths.  (I think possibly I disregarded that in
the original coding on the theory that nested appendrels would get folded
together --- but that only happens *after* reparameterize_path(), so it's
not excused from handling a child AppendPath.)  Given that this code's been
like this since 9.3 when LATERAL was introduced, it seems likely we'd have
heard of other cases by now if there were a larger problem.

Per report from Elvis Pranskevichus.  Back-patch to 9.3.

Discussion: https://postgr.es/m/5981018.zdth1YWmNy@hammer.magicstack.net

diff --git a/src/backend/optimizer/util/pathnode.c b/src/backend/optimizer/util/pathnode.c
index f2d6385f18353ad4e4bf93e2db1c26ac493b6c25..b34c5f869005804df1c209bdfb41bf2960b6eb00 100644 (file)
--- a/src/backend/optimizer/util/pathnode.c
+++ b/src/backend/optimizer/util/pathnode.c
@@ -3423,6 +3423,30 @@ reparameterize_path(PlannerInfo *root, Path *path,
                                                         spath->path.pathkeys,
                                                         required_outer);
            }
+       case T_Append:
+           {
+               AppendPath *apath = (AppendPath *) path;
+               List       *childpaths = NIL;
+               ListCell   *lc;
+
+               /* Reparameterize the children */
+               foreach(lc, apath->subpaths)
+               {
+                   Path       *spath = (Path *) lfirst(lc);
+
+                   spath = reparameterize_path(root, spath,
+                                               required_outer,
+                                               loop_count);
+                   if (spath == NULL)
+                       return NULL;
+                   childpaths = lappend(childpaths, spath);
+               }
+               return (Path *)
+                   create_append_path(rel, childpaths,
+                                      required_outer,
+                                      apath->path.parallel_workers,
+                                      apath->partitioned_rels);
+           }
        default:
            break;
    }

diff --git a/src/test/regress/expected/join.out b/src/test/regress/expected/join.out
index 9f4c88dab4ed76bdffe76a9687a44b872d01577e..b007ffd98d3af56de1c8d25b0926696a3e621194 100644 (file)
--- a/src/test/regress/expected/join.out
+++ b/src/test/regress/expected/join.out
@@ -5154,6 +5154,25 @@ select * from
          Output: 3
 (11 rows)
 
+-- check handling of nested appendrels inside LATERAL
+select * from
+  ((select 2 as v) union all (select 3 as v)) as q1
+  cross join lateral
+  ((select * from
+      ((select 4 as v) union all (select 5 as v)) as q3)
+   union all
+   (select q1.v)
+  ) as q2;
+ v | v 
+---+---
+ 2 | 4
+ 2 | 5
+ 2 | 2
+ 3 | 4
+ 3 | 5
+ 3 | 3
+(6 rows)
+
 -- check we don't try to do a unique-ified semijoin with LATERAL
 explain (verbose, costs off)
 select * from

diff --git a/src/test/regress/sql/join.sql b/src/test/regress/sql/join.sql
index 835d67551cd56efed7f72f2575b7ad56c2a10ef0..fa05b0b9b0fb531e79e9f7800e1763754e5f7ec7 100644 (file)
--- a/src/test/regress/sql/join.sql
+++ b/src/test/regress/sql/join.sql
@@ -1668,6 +1668,16 @@ select * from
     select * from (select 3 as z offset 0) z where z.z = x.x
   ) zz on zz.z = y.y;
 
+-- check handling of nested appendrels inside LATERAL
+select * from
+  ((select 2 as v) union all (select 3 as v)) as q1
+  cross join lateral
+  ((select * from
+      ((select 4 as v) union all (select 5 as v)) as q3)
+   union all
+   (select q1.v)
+  ) as q2;
+
 -- check we don't try to do a unique-ified semijoin with LATERAL
 explain (verbose, costs off)
 select * from