The following bug has been logged online:
Bug reference: 3308
Logged by: George Sakkis
Email address: [EMAIL PROTECTED]
PostgreSQL version: 8.2.3
Operating system: Linux
Description:Nested outer join gives wrong results
Details:
I have the following two subqueries, both returning a single column q_id;
also the result set of A is a strict superset of B:
Query A
---
SELECT query.id as q_id
FROM ranker, run, query
WHERE ranker.id = 72 AND
run.id = ranker.run_id AND
query.set_id = run.set_id
(966 rows)
Query B
---
SELECT serp_result.q_id
FROM serp_result LEFT OUTER JOIN editor_rating using (q_id,norm_url)
WHERE serp_result.ranker_id = 72 AND
serp_result.rank <= 1
AND editor_rating.grade is null
(251 rows)
Now, the left outer join of A and B should be equal to A in this case since
A is a superset of B. If I save A in a temp table and use this for the join,
that's indeed the result:
SELECT query.id as q_id INTO TEMP t1
FROM ranker, run, query
WHERE ranker.id = 72 AND
run.id = ranker.run_id AND
query.set_id = run.set_id
SELECT *
FROM t1
LEFT JOIN (
SELECT serp_result.q_id
FROM serp_result LEFT OUTER JOIN editor_rating using (q_id,norm_url)
WHERE serp_result.ranker_id = 72 AND serp_result.rank <= 1 AND
editor_rating.grade is null
) AS t2 USING (q_id)
;
(966 rows)
Likewise if I save B into a temp table and join with A:
SELECT serp_result.q_id into TEMP t2
FROM serp_result LEFT OUTER JOIN editor_rating using (q_id,norm_url)
WHERE serp_result.ranker_id = 72 AND serp_result.rank <= 1 AND
editor_rating.grade is null
SELECT *
FROM (
SELECT query.id as q_id
FROM ranker, run, query
WHERE ranker.id = 72 AND
run.id = ranker.run_id AND
query.set_id = run.set_id
) AS t1
LEFT JOIN t2 USING (q_id)
(966 rows)
If I don't use temp tables though, the result is equal to B instead:
SELECT *
FROM (
SELECT query.id as q_id
FROM ranker, run, query
WHERE ranker.id = 72 AND
run.id = ranker.run_id AND
query.set_id = run.set_id
) AS t1
LEFT JOIN (
SELECT serp_result.q_id
FROM serp_result LEFT OUTER JOIN editor_rating using (q_id,norm_url)
WHERE serp_result.ranker_id = 72 AND
serp_result.rank <= 1
AND editor_rating.grade is null
) AS t2 USING (q_id)
;
(251 rows)
What gives ?
---(end of broadcast)---
TIP 9: In versions below 8.0, the planner will ignore your desire to
choose an index scan if your joining column's datatypes do not
match
