假设我有表student,club和student_club:
student {
id
name
}
club {
id
name
}
student_club {
student_id
club_id
}
我想知道如何找到足球 (30) 和棒球 (50) 俱乐部的所有学生。
虽然这个查询不起作用，但它是我迄今为止最接近的东西：
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
我很好奇。众所周知，好奇心以杀死猫而闻名。
本次测试的猫皮环境：
student.idis student.stud_idand club.idis club.club_idhere。EXPLAIN ANALYZE。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_pkey这里的大多数查询都不需要。
主键在 PostgreSQL 中自动实现唯一索引。
最后一个索引是为了弥补PostgreSQL上多列索引的这个已知缺点：
多列 B 树索引可用于涉及索引列的任何子集的查询条件，但当前导（最左侧）列存在约束时，索引效率最高。
总运行时间EXPLAIN ANALYZE。
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;
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);
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
);
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);
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);
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;
最后三个的表现几乎相同。4) 和 5) 产生相同的查询计划。
花哨的 SQL，但性能跟不上：
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
)
);
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
)
);
正如预期的那样，这两者的表现几乎相同。查询计划导致表扫描，计划程序在这里找不到使用索引的方法。
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;
花哨的 SQL，CTE 的良好性能。非常奇特的查询计划。
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);
查询 2) 的 CTE 变体。令人惊讶的是，它可能会导致使用完全相同的数据的查询计划略有不同。我在 上找到了顺序扫描student，其中子查询变体使用了索引。
另一个后期添加的超级立方体。真是太神奇了，有多少种方法。
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
)
);
ypercube 的 11) 实际上只是这个更简单的变体的令人费解的反向方法，它仍然缺失。执行速度几乎与顶级猫科动物一样快。
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
);
难以置信，但这是另一个真正的新变体。我看到了超过两个会员资格的潜力，但它也跻身于只有两个会员的顶级猫之列。
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
);
换句话说：不同数量的过滤器。这个问题要求恰好有两个俱乐部会员资格。但是许多用例必须为不同的数量做准备。看：
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
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)
如果你只想要 student_id 那么：
Select student_id
from student_club
where club_id in ( 30, 50 )
group by student_id
having count( student_id ) = 2
如果您还需要学生的姓名，那么：
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 )
如果您在一个 club_selection 表中有两个以上的俱乐部，那么：
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 ) )
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)
或者更通用的解决方案更容易扩展到n俱乐部并且避免INTERSECT（在 MySQL 中不可用）和IN（因为它的性能在 MySQL 中很糟糕）
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
所以给猫剥皮的方法不止一种。
我将再添加两个以使其更完整。
假设一个健全的(student_id, club_id)数据模型在student_club. Martin Smith 的第二个版本有点类似，但他先加入，然后再分组。这应该更快：
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);
当然，还有经典EXISTS。类似于 Derek 的变体，带有IN. 简单快速。（在 MySQL 中，这应该比带有 的变体快很多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);
另一个 CTE。它看起来很干净，但它可能会生成与普通子查询中的 groupby 相同的计划。
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)
;
对于那些想要测试的人，我的生成测试数据的副本：
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)
;
由于没有人添加此（经典）版本：
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
)
)
或类似的：
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
)
)
再尝试一种稍微不同的方法。受解释扩展中的一篇文章的启发：EAV 表中的多个属性： GROUP BY 与 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
)
)
另一种方法：
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 ;
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
;
这似乎表现得相当好，因为 CTE 扫描避免了对两个单独的子查询的需要。
滥用递归查询总是有原因的！
（顺便说一句：mysql似乎没有递归查询）
我在现实生活中的数据库中进行了测试，因此名称与猫皮列表不同。这是一个备份副本，因此在所有测试运行期间没有任何变化（除了对目录的微小更改）。
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
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
@erwin-brandstetter 请对此进行基准测试：
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;
这就像@sean 的 6 号），我猜是更干净。
-- 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)
;
查询计划：
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)
所以它似乎仍然想要对学生进行 seq 扫描。
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
;
使用最快的变体（Brandstetter 先生图表中的 Sean 先生）。可能只有一个加入到只有 student_club 矩阵有生存权的变体。因此，最长的查询将只有两列要计算，想法是使查询变细。