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Fix usage of the parse tree for estimate_num_groups() in set operations
recurse_set_operations() uses the parse tree for the group number estimation, because of the "varno 0" hack. At the same time 2489d76c49 made root->parse and corresponding parent_root->simple_rte_array[]->subquery distinct copies of the parse tree, while d3d55ce571 introduced self-join removal replacing relid of removed relation only in one of the copies. The present commit fixes this bug by making recurse_set_operations() call estimate_num_groups() with the copy of the parse tree processed by self-join removal. In future, we may think about maintaining just one copy of the parse tree and/or keeping removed relids as aliases. Reported-by: Zuming Jiang Bug: #18170 Discussion: https://postgr.es/m/flat/18170-f1d17bf9a0d58b24%40postgresql.org Author: Richard Guo, Alexander Korotkov Reviewed-by: Andrei Lepikhov
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@ -324,7 +324,7 @@ recurse_set_operations(Node *setOp, PlannerInfo *root,
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* anyway; otherwise do statistical estimation.
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* anyway; otherwise do statistical estimation.
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*
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*
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* XXX you don't really want to know about this: we do the estimation
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* XXX you don't really want to know about this: we do the estimation
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* using the subquery's original targetlist expressions, not the
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* using the subroot->parse's original targetlist expressions, not the
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* subroot->processed_tlist which might seem more appropriate. The
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* subroot->processed_tlist which might seem more appropriate. The
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* reason is that if the subquery is itself a setop, it may return a
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* reason is that if the subquery is itself a setop, it may return a
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* processed_tlist containing "varno 0" Vars generated by
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* processed_tlist containing "varno 0" Vars generated by
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@ -332,6 +332,9 @@ recurse_set_operations(Node *setOp, PlannerInfo *root,
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* mightily. We ought to get rid of the "varno 0" hack, but that
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* mightily. We ought to get rid of the "varno 0" hack, but that
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* requires a redesign of the parsetree representation of setops, so
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* requires a redesign of the parsetree representation of setops, so
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* that there can be an RTE corresponding to each setop's output.
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* that there can be an RTE corresponding to each setop's output.
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* Note, we use this not subquery's targetlist but subroot->parse's
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* targetlist, because it was revised by self-join removal. subquery's
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* targetlist might contain the references to the removed relids.
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*/
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*/
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if (pNumGroups)
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if (pNumGroups)
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{
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{
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@ -341,7 +344,7 @@ recurse_set_operations(Node *setOp, PlannerInfo *root,
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*pNumGroups = subpath->rows;
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*pNumGroups = subpath->rows;
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else
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else
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*pNumGroups = estimate_num_groups(subroot,
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*pNumGroups = estimate_num_groups(subroot,
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get_tlist_exprs(subquery->targetList, false),
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get_tlist_exprs(subroot->parse->targetList, false),
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subpath->rows,
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subpath->rows,
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NULL,
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NULL,
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NULL);
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NULL);
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@ -6759,6 +6759,23 @@ WHERE c3.id=c2.id AND c3.id*c2.id=c1.id*c1.id;
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Filter: ((id IS NOT NULL) AND ((id * id) IS NOT NULL))
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Filter: ((id IS NOT NULL) AND ((id * id) IS NOT NULL))
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(3 rows)
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(3 rows)
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-- Check the usage of a parse tree by the set operations (bug #18170)
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explain (costs off)
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SELECT c1.code FROM emp1 c1 LEFT JOIN emp1 c2 ON c1.id = c2.id
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WHERE c2.id IS NOT NULL
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EXCEPT ALL
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SELECT c3.code FROM emp1 c3;
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QUERY PLAN
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----------------------------------------------
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HashSetOp Except All
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-> Append
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-> Subquery Scan on "*SELECT* 1"
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-> Seq Scan on emp1 c2
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Filter: (id IS NOT NULL)
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-> Subquery Scan on "*SELECT* 2"
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-> Seq Scan on emp1 c3
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(7 rows)
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-- We can remove the join even if we find the join can't duplicate rows and
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-- We can remove the join even if we find the join can't duplicate rows and
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-- the base quals of each side are different. In the following case we end up
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-- the base quals of each side are different. In the following case we end up
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-- moving quals over to s1 to make it so it can't match any rows.
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-- moving quals over to s1 to make it so it can't match any rows.
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@ -2573,6 +2573,13 @@ explain (costs off)
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SELECT count(*) FROM emp1 c1, emp1 c2, emp1 c3
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SELECT count(*) FROM emp1 c1, emp1 c2, emp1 c3
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WHERE c3.id=c2.id AND c3.id*c2.id=c1.id*c1.id;
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WHERE c3.id=c2.id AND c3.id*c2.id=c1.id*c1.id;
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-- Check the usage of a parse tree by the set operations (bug #18170)
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explain (costs off)
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SELECT c1.code FROM emp1 c1 LEFT JOIN emp1 c2 ON c1.id = c2.id
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WHERE c2.id IS NOT NULL
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EXCEPT ALL
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SELECT c3.code FROM emp1 c3;
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-- We can remove the join even if we find the join can't duplicate rows and
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-- We can remove the join even if we find the join can't duplicate rows and
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-- the base quals of each side are different. In the following case we end up
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-- the base quals of each side are different. In the following case we end up
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-- moving quals over to s1 to make it so it can't match any rows.
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-- moving quals over to s1 to make it so it can't match any rows.
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