Re: [HACKERS] [GENERAL] Recursive optimization of IN subqueries

[Dennis Haney]

Tom Lane wrote: Dennis Haney <[EMAIL PROTECTED]> writes: I saw it as though convert_IN_to_join rewrote the query from select a.* from tenk1 a where a.unique1 in (select c.thousand from tenk1 c where c.hundred = 99); to select a.* from tenk1 a, tenk1 c where a.unique1 = c.thousand AND c.hundred = 99; but after looking at it, I've reached the conclusion that the rewrite is to this instead: select a.* from tenk1 a, (select d.thousand from tenk1 d where d.hundred = 99) as c where a.unique1 = c.thousand; Right. We do that, and then subsequently pull_up_subqueries transforms it to the other representation. The reason for this two-step approach is that the intermediate form is still a useful improvement if the subquery can't be pulled up for some reason (e.g., it's got grouping). With improvement I can see that it can be materialized and thus used as a normal table in the planner. Is there any additional reasons that I can't see? But this limited optimization makes me wonder, why the limitation to optimizing '='? And why must the lefthand of the sublink be a variable of the upper query? except the subselect is added as a range table entry instead of a subselect in the from-list (not that I understand this particular part, do you mind explaining?). Same thing. Every entry in the from-list will have both an RTE and an entry in the join tree. This representation is partly historical (before we had outer joins, there was only the range table and no join tree at all), but it is convenient for many purposes. Then I don't understand why it gives two different execution plans? And the Query* is totally different for the two, eg. there is no RTE for the subquery in the first query: davh=# explain select a.* from test1 a, (select num from test1 where id = 2) as b where a.num = b.num; QUERY PLAN ------------------------------------------------------------------------------------ Hash Join (cost=4.83..29.94 rows=11 width=8) Hash Cond: ("outer".num = "inner".num) -> Seq Scan on test1 a (cost=0.00..20.00 rows=1000 width=8) -> Hash (cost=4.82..4.82 rows=2 width=4) -> Index Scan using test1_pkey on test1 (cost=0.00..4.82 rows=2 width=4) Index Cond: (id = 2) (6 rows) davh=# explain select a.* from test1 a where a.num in (select num from test1 where id = 2); QUERY PLAN ------------------------------------------------------------------------------------ Hash IN Join (cost=4.83..28.75 rows=6 width=8) Hash Cond: ("outer".num = "inner".num) -> Seq Scan on test1 a (cost=0.00..20.00 rows=1000 width=8) -> Hash (cost=4.82..4.82 rows=2 width=4) -> Index Scan using test1_pkey on test1 (cost=0.00..4.82 rows=2 width=4) Index Cond: (id = 2) (6 rows) PS: this is a bit off-topic for pgsql-general, please pursue it on -hackers if you have more questions.ok -- Dennis