问题
Assuming I have the tables student, club, and student_club:
student {
id
name
}
club {
id
name
}
student_club {
student_id
club_id
}
I want to know how to find all students in both the soccer (30) and baseball (50) club.
While this query doesn\'t work, it\'s the closest thing I have so far:
SELECT student.*
FROM student
INNER JOIN student_club sc ON student.id = sc.student_id
LEFT JOIN club c ON c.id = sc.club_id
WHERE c.id = 30 AND c.id = 50
回答1:
I was curious. And as we all know, curiosity has a reputation for killing cats.
So, which is the fastest way to skin a cat?
The precise cat-skinning environment for this test:
- PostgreSQL 9.0 on Debian Squeeze with decent RAM and settings.
- 6.000 students, 24.000 club memberships (data copied from a similar database with real life data.)
- Slight diversion from the naming schema in the question:
student.idisstudent.stud_idandclub.idisclub.club_idhere. - I named the queries after their author in this thread, with an index where there are two.
- I ran all queries a couple of times to populate the cache, then I picked the best of 5 with EXPLAIN ANALYZE.
Relevant indexes (should be the optimum - as long as we lack fore-knowledge which clubs will be queried):
ALTER TABLE student ADD CONSTRAINT student_pkey PRIMARY KEY(stud_id ); ALTER TABLE student_club ADD CONSTRAINT sc_pkey PRIMARY KEY(stud_id, club_id); ALTER TABLE club ADD CONSTRAINT club_pkey PRIMARY KEY(club_id ); CREATE INDEX sc_club_id_idx ON student_club (club_id);club_pkeyis not required by most queries here.
Primary keys implement unique indexes automatically In PostgreSQL.
The last index is to make up for this known shortcoming of multi-column indexes on PostgreSQL:
A multicolumn B-tree index can be used with query conditions that involve any subset of the index's columns, but the index is most efficient when there are constraints on the leading (leftmost) columns.
Results:
Total runtimes from EXPLAIN ANALYZE.
1) Martin 2: 44.594 ms
SELECT s.stud_id, s.name
FROM student s
JOIN student_club sc USING (stud_id)
WHERE sc.club_id IN (30, 50)
GROUP BY 1,2
HAVING COUNT(*) > 1;
2) Erwin 1: 33.217 ms
SELECT s.stud_id, s.name
FROM student s
JOIN (
SELECT stud_id
FROM student_club
WHERE club_id IN (30, 50)
GROUP BY 1
HAVING COUNT(*) > 1
) sc USING (stud_id);
3) Martin 1: 31.735 ms
SELECT s.stud_id, s.name
FROM student s
WHERE student_id IN (
SELECT student_id
FROM student_club
WHERE club_id = 30
INTERSECT
SELECT stud_id
FROM student_club
WHERE club_id = 50);
4) Derek: 2.287 ms
SELECT s.stud_id, s.name
FROM student s
WHERE s.stud_id IN (SELECT stud_id FROM student_club WHERE club_id = 30)
AND s.stud_id IN (SELECT stud_id FROM student_club WHERE club_id = 50);
5) Erwin 2: 2.181 ms
SELECT s.stud_id, s.name
FROM student s
WHERE EXISTS (SELECT * FROM student_club
WHERE stud_id = s.stud_id AND club_id = 30)
AND EXISTS (SELECT * FROM student_club
WHERE stud_id = s.stud_id AND club_id = 50);
6) Sean: 2.043 ms
SELECT s.stud_id, s.name
FROM student s
JOIN student_club x ON s.stud_id = x.stud_id
JOIN student_club y ON s.stud_id = y.stud_id
WHERE x.club_id = 30
AND y.club_id = 50;
The last three perform pretty much the same. 4) and 5) result in the same query plan.
Late Additions:
Fancy SQL, but the performance can't keep up.
7) ypercube 1: 148.649 ms
SELECT s.stud_id, s.name
FROM student AS s
WHERE NOT EXISTS (
SELECT *
FROM club AS c
WHERE c.club_id IN (30, 50)
AND NOT EXISTS (
SELECT *
FROM student_club AS sc
WHERE sc.stud_id = s.stud_id
AND sc.club_id = c.club_id
)
);
8) ypercube 2: 147.497 ms
SELECT s.stud_id, s.name
FROM student AS s
WHERE NOT EXISTS (
SELECT *
FROM (
SELECT 30 AS club_id
UNION ALL
SELECT 50
) AS c
WHERE NOT EXISTS (
SELECT *
FROM student_club AS sc
WHERE sc.stud_id = s.stud_id
AND sc.club_id = c.club_id
)
);
As expected, those two perform almost the same. Query plan results in table scans, the planner doesn't find a way to use the indexes here.
9) wildplasser 1: 49.849 ms
WITH RECURSIVE two AS (
SELECT 1::int AS level
, stud_id
FROM student_club sc1
WHERE sc1.club_id = 30
UNION
SELECT two.level + 1 AS level
, sc2.stud_id
FROM student_club sc2
JOIN two USING (stud_id)
WHERE sc2.club_id = 50
AND two.level = 1
)
SELECT s.stud_id, s.student
FROM student s
JOIN two USING (studid)
WHERE two.level > 1;
Fancy SQL, decent performance for a CTE. Very exotic query plan.
Again, would be interesting how 9.1 handles this. I am going to upgrade the db cluster used here to 9.1 soon. Maybe I'll rerun the whole shebang ...
10) wildplasser 2: 36.986 ms
WITH sc AS (
SELECT stud_id
FROM student_club
WHERE club_id IN (30,50)
GROUP BY stud_id
HAVING COUNT(*) > 1
)
SELECT s.*
FROM student s
JOIN sc USING (stud_id);
CTE variant of query 2). Surprisingly, it can result in a slightly different query plan with the exact same data. I found a sequential scan on student, where the subquery-variant used the index.
11) ypercube 3: 101.482 ms
Another late addition @ypercube. It is positively amazing, how many ways there are.
SELECT s.stud_id, s.student
FROM student s
JOIN student_club sc USING (stud_id)
WHERE sc.club_id = 10 -- member in 1st club ...
AND NOT EXISTS (
SELECT *
FROM (SELECT 14 AS club_id) AS c -- can't be excluded for missing the 2nd
WHERE NOT EXISTS (
SELECT *
FROM student_club AS d
WHERE d.stud_id = sc.stud_id
AND d.club_id = c.club_id
)
)
12) erwin 3: 2.377 ms
@ypercube's 11) is actually just the mind-twisting reverse approach of this simpler variant, that was also still missing. Performs almost as fast as the top cats.
SELECT s.*
FROM student s
JOIN student_club x USING (stud_id)
WHERE sc.club_id = 10 -- member in 1st club ...
AND EXISTS ( -- ... and membership in 2nd exists
SELECT *
FROM student_club AS y
WHERE y.stud_id = s.stud_id
AND y.club_id = 14
)
13) erwin 4: 2.375 ms
Hard to believe, but here's another, genuinely new variant. I see potential for more than two memberships, but it also ranks among the top cats with just two.
SELECT s.*
FROM student AS s
WHERE EXISTS (
SELECT *
FROM student_club AS x
JOIN student_club AS y USING (stud_id)
WHERE x.stud_id = s.stud_id
AND x.club_id = 14
AND y.club_id = 10
)
Dynamic number of club memberships
In other words: varying number of filters. This question asked for exactly two club memberships. But many use cases have to prepare for a varying number.
Detailed discussion in this related later answer:
- Using same column multiple times in WHERE clause
回答2:
SELECT s.*
FROM student s
INNER JOIN student_club sc_soccer ON s.id = sc_soccer.student_id
INNER JOIN student_club sc_baseball ON s.id = sc_baseball.student_id
WHERE
sc_baseball.club_id = 50 AND
sc_soccer.club_id = 30
回答3:
select *
from student
where id in (select student_id from student_club where club_id = 30)
and id in (select student_id from student_club where club_id = 50)
回答4:
If you just want student_id then:
Select student_id
from student_club
where club_id in ( 30, 50 )
group by student_id
having count( student_id ) = 2
If you also need name from student then:
Select student_id, name
from student s
where exists( select *
from student_club sc
where s.student_id = sc.student_id
and club_id in ( 30, 50 )
group by sc.student_id
having count( sc.student_id ) = 2 )
If you have more than two clubs in a club_selection table then:
Select student_id, name
from student s
where exists( select *
from student_club sc
where s.student_id = sc.student_id
and exists( select *
from club_selection cs
where sc.club_id = cs.club_id )
group by sc.student_id
having count( sc.student_id ) = ( select count( * )
from club_selection ) )
回答5:
SELECT *
FROM student
WHERE id IN (SELECT student_id
FROM student_club
WHERE club_id = 30
INTERSECT
SELECT student_id
FROM student_club
WHERE club_id = 50)
Or a more general solution easier to extend to n clubs and that avoids INTERSECT (not available in MySQL) and IN (as performance of this sucks in MySQL)
SELECT s.id,
s.name
FROM student s
join student_club sc
ON s.id = sc.student_id
WHERE sc.club_id IN ( 30, 50 )
GROUP BY s.id,
s.name
HAVING COUNT(DISTINCT sc.club_id) = 2
回答6:
Another CTE. It looks clean, but it will probably generate the same plan as a groupby in a normal subquery.
WITH two AS (
SELECT student_id FROM tmp.student_club
WHERE club_id IN (30,50)
GROUP BY student_id
HAVING COUNT(*) > 1
)
SELECT st.* FROM tmp.student st
JOIN two ON (two.student_id=st.id)
;
For those who want to test, a copy of my generate testdata thingy:
DROP SCHEMA tmp CASCADE;
CREATE SCHEMA tmp;
CREATE TABLE tmp.student
( id INTEGER NOT NULL PRIMARY KEY
, sname VARCHAR
);
CREATE TABLE tmp.club
( id INTEGER NOT NULL PRIMARY KEY
, cname VARCHAR
);
CREATE TABLE tmp.student_club
( student_id INTEGER NOT NULL REFERENCES tmp.student(id)
, club_id INTEGER NOT NULL REFERENCES tmp.club(id)
);
INSERT INTO tmp.student(id)
SELECT generate_series(1,1000)
;
INSERT INTO tmp.club(id)
SELECT generate_series(1,100)
;
INSERT INTO tmp.student_club(student_id,club_id)
SELECT st.id , cl.id
FROM tmp.student st, tmp.club cl
;
DELETE FROM tmp.student_club
WHERE random() < 0.8
;
UPDATE tmp.student SET sname = 'Student#' || id::text ;
UPDATE tmp.club SET cname = 'Soccer' WHERE id = 30;
UPDATE tmp.club SET cname = 'Baseball' WHERE id = 50;
ALTER TABLE tmp.student_club
ADD PRIMARY KEY (student_id,club_id)
;
回答7:
So there's more than one way to skin a cat.
I'll to add two more to make it, well, more complete.
1) GROUP first, JOIN later
Assuming a sane data model where (student_id, club_id) is unique in student_club.
Martin Smith's second version is like somewhat similar, but he joins first, groups later. This should be faster:
SELECT s.id, s.name
FROM student s
JOIN (
SELECT student_id
FROM student_club
WHERE club_id IN (30, 50)
GROUP BY 1
HAVING COUNT(*) > 1
) sc USING (student_id);
2) EXISTS
And of course, there is the classic EXISTS. Similar to Derek's variant with IN. Simple and fast. (In MySQL, this should be quite a bit faster than the variant with IN):
SELECT s.id, s.name
FROM student s
WHERE EXISTS (SELECT 1 FROM student_club
WHERE student_id = s.student_id AND club_id = 30)
AND EXISTS (SELECT 1 FROM student_club
WHERE student_id = s.student_id AND club_id = 50);
回答8:
Since noone has added this (classic) version:
SELECT s.*
FROM student AS s
WHERE NOT EXISTS
( SELECT *
FROM club AS c
WHERE c.id IN (30, 50)
AND NOT EXISTS
( SELECT *
FROM student_club AS sc
WHERE sc.student_id = s.id
AND sc.club_id = c.id
)
)
or similar:
SELECT s.*
FROM student AS s
WHERE NOT EXISTS
( SELECT *
FROM
( SELECT 30 AS club_id
UNION ALL
SELECT 50
) AS c
WHERE NOT EXISTS
( SELECT *
FROM student_club AS sc
WHERE sc.student_id = s.id
AND sc.club_id = c.club_id
)
)
One more try with a slightly different approach. Inspired by an article in Explain Extended: Multiple attributes in a EAV table: GROUP BY vs. NOT EXISTS:
SELECT s.*
FROM student_club AS sc
JOIN student AS s
ON s.student_id = sc.student_id
WHERE sc.club_id = 50 --- one option here
AND NOT EXISTS
( SELECT *
FROM
( SELECT 30 AS club_id --- all the rest in here
--- as in previous query
) AS c
WHERE NOT EXISTS
( SELECT *
FROM student_club AS scc
WHERE scc.student_id = sc.id
AND scc.club_id = c.club_id
)
)
Another approach:
SELECT s.stud_id
FROM student s
EXCEPT
SELECT stud_id
FROM
( SELECT s.stud_id, c.club_id
FROM student s
CROSS JOIN (VALUES (30),(50)) c (club_id)
EXCEPT
SELECT stud_id, club_id
FROM student_club
WHERE club_id IN (30, 50) -- optional. Not needed but may affect performance
) x ;
回答9:
WITH RECURSIVE two AS
( SELECT 1::integer AS level
, student_id
FROM tmp.student_club sc0
WHERE sc0.club_id = 30
UNION
SELECT 1+two.level AS level
, sc1.student_id
FROM tmp.student_club sc1
JOIN two ON (two.student_id = sc1.student_id)
WHERE sc1.club_id = 50
AND two.level=1
)
SELECT st.* FROM tmp.student st
JOIN two ON (two.student_id=st.id)
WHERE two.level> 1
;
This seems to perform reasonably well, since the CTE-scan avoids the need for two separate subqueries.
There is always a reason to misuse recursive queries!
(BTW: mysql does not seem to have recursive queries)
回答10:
Different query plans in query 2) and 10)
I tested in a real life db, so the names differ from the catskin list. It's a backup copy, so nothing changed during all test runs (except minor changes to the catalogs).
Query 2)
SELECT a.*
FROM ef.adr a
JOIN (
SELECT adr_id
FROM ef.adratt
WHERE att_id IN (10,14)
GROUP BY adr_id
HAVING COUNT(*) > 1) t using (adr_id);
Merge Join (cost=630.10..1248.78 rows=627 width=295) (actual time=13.025..34.726 rows=67 loops=1)
Merge Cond: (a.adr_id = adratt.adr_id)
-> Index Scan using adr_pkey on adr a (cost=0.00..523.39 rows=5767 width=295) (actual time=0.023..11.308 rows=5356 loops=1)
-> Sort (cost=630.10..636.37 rows=627 width=4) (actual time=12.891..13.004 rows=67 loops=1)
Sort Key: adratt.adr_id
Sort Method: quicksort Memory: 28kB
-> HashAggregate (cost=450.87..488.49 rows=627 width=4) (actual time=12.386..12.710 rows=67 loops=1)
Filter: (count(*) > 1)
-> Bitmap Heap Scan on adratt (cost=97.66..394.81 rows=2803 width=4) (actual time=0.245..5.958 rows=2811 loops=1)
Recheck Cond: (att_id = ANY ('{10,14}'::integer[]))
-> Bitmap Index Scan on adratt_att_id_idx (cost=0.00..94.86 rows=2803 width=0) (actual time=0.217..0.217 rows=2811 loops=1)
Index Cond: (att_id = ANY ('{10,14}'::integer[]))
Total runtime: 34.928 ms
Query 10)
WITH two AS (
SELECT adr_id
FROM ef.adratt
WHERE att_id IN (10,14)
GROUP BY adr_id
HAVING COUNT(*) > 1
)
SELECT a.*
FROM ef.adr a
JOIN two using (adr_id);
Hash Join (cost=1161.52..1261.84 rows=627 width=295) (actual time=36.188..37.269 rows=67 loops=1)
Hash Cond: (two.adr_id = a.adr_id)
CTE two
-> HashAggregate (cost=450.87..488.49 rows=627 width=4) (actual time=13.059..13.447 rows=67 loops=1)
Filter: (count(*) > 1)
-> Bitmap Heap Scan on adratt (cost=97.66..394.81 rows=2803 width=4) (actual time=0.252..6.252 rows=2811 loops=1)
Recheck Cond: (att_id = ANY ('{10,14}'::integer[]))
-> Bitmap Index Scan on adratt_att_id_idx (cost=0.00..94.86 rows=2803 width=0) (actual time=0.226..0.226 rows=2811 loops=1)
Index Cond: (att_id = ANY ('{10,14}'::integer[]))
-> CTE Scan on two (cost=0.00..50.16 rows=627 width=4) (actual time=13.065..13.677 rows=67 loops=1)
-> Hash (cost=384.68..384.68 rows=5767 width=295) (actual time=23.097..23.097 rows=5767 loops=1)
Buckets: 1024 Batches: 1 Memory Usage: 1153kB
-> Seq Scan on adr a (cost=0.00..384.68 rows=5767 width=295) (actual time=0.005..10.955 rows=5767 loops=1)
Total runtime: 37.482 ms
回答11:
@erwin-brandstetter Please, benchmark this:
SELECT s.stud_id, s.name
FROM student s, student_club x, student_club y
WHERE x.club_id = 30
AND s.stud_id = x.stud_id
AND y.club_id = 50
AND s.stud_id = y.stud_id;
It's like number 6) by @sean , just cleaner, I guess.
回答12:
-- EXPLAIN ANALYZE
WITH two AS (
SELECT c0.student_id
FROM tmp.student_club c0
, tmp.student_club c1
WHERE c0.student_id = c1.student_id
AND c0.club_id = 30
AND c1.club_id = 50
)
SELECT st.* FROM tmp.student st
JOIN two ON (two.student_id=st.id)
;
The query plan:
Hash Join (cost=1904.76..1919.09 rows=337 width=15) (actual time=6.937..8.771 rows=324 loops=1)
Hash Cond: (two.student_id = st.id)
CTE two
-> Hash Join (cost=849.97..1645.76 rows=337 width=4) (actual time=4.932..6.488 rows=324 loops=1)
Hash Cond: (c1.student_id = c0.student_id)
-> Bitmap Heap Scan on student_club c1 (cost=32.76..796.94 rows=1614 width=4) (actual time=0.667..1.835 rows=1646 loops=1)
Recheck Cond: (club_id = 50)
-> Bitmap Index Scan on sc_club_id_idx (cost=0.00..32.36 rows=1614 width=0) (actual time=0.473..0.473 rows=1646 loops=1)
Index Cond: (club_id = 50)
-> Hash (cost=797.00..797.00 rows=1617 width=4) (actual time=4.203..4.203 rows=1620 loops=1)
Buckets: 1024 Batches: 1 Memory Usage: 57kB
-> Bitmap Heap Scan on student_club c0 (cost=32.79..797.00 rows=1617 width=4) (actual time=0.663..3.596 rows=1620 loops=1)
Recheck Cond: (club_id = 30)
-> Bitmap Index Scan on sc_club_id_idx (cost=0.00..32.38 rows=1617 width=0) (actual time=0.469..0.469 rows=1620 loops=1)
Index Cond: (club_id = 30)
-> CTE Scan on two (cost=0.00..6.74 rows=337 width=4) (actual time=4.935..6.591 rows=324 loops=1)
-> Hash (cost=159.00..159.00 rows=8000 width=15) (actual time=1.979..1.979 rows=8000 loops=1)
Buckets: 1024 Batches: 1 Memory Usage: 374kB
-> Seq Scan on student st (cost=0.00..159.00 rows=8000 width=15) (actual time=0.093..0.759 rows=8000 loops=1)
Total runtime: 8.989 ms
(20 rows)
So it still seems to want the seq scan on student.
回答13:
SELECT s.stud_id, s.name
FROM student s,
(
select x.stud_id from
student_club x
JOIN student_club y ON x.stud_id = y.stud_id
WHERE x.club_id = 30
AND y.club_id = 50
) tmp_tbl
where tmp_tbl.stud_id = s.stud_id
;
Use of fastest variant (Mr. Sean in Mr. Brandstetter chart). May be variant with only one join to only the student_club matrix has the right to live. So, the longest query will have only two columns to calculate, idea is to make the query thin.
来源：https://stackoverflow.com/questions/7364969/how-to-filter-sql-results-in-a-has-many-through-relation