如何对SQL结果进行过滤，使其具有一定的相关性

Assuming I have the tables student, club, and student_club:

假设我有学生，俱乐部和学生俱乐部:

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:

我想知道如何在足球(30)和棒球(50)俱乐部找到所有的学生。虽然这个查询不起作用，但它是我到目前为止最接近的东西:

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

13 个解决方案

#1

I was curious. And as we all know, curiosity has a reputation for killing cats.

我很好奇。我们都知道，好奇心是杀死猫的好名声。

So, which is the fastest way to skin a cat?

The precise cat-skinning environment for this test:

本次试验的精确剥猫环境:

• PostgreSQL 9.0 on Debian Squeeze with decent RAM and settings.
• Debian压缩包上的PostgreSQL 9.0具有良好的RAM和设置。
• 6.000 students, 24.000 club memberships (data copied from a similar database with real life data.)
• 6.000名学生，24.000名俱乐部会员(数据来自真实生活数据的类似数据库)。
• Slight diversion from the naming schema in the question: student.id is student.stud_id and club.id is club.club_id here.
• 稍微偏离题目中的命名模式:学生。id是学生。stud_id和俱乐部。id是俱乐部。club_id这里。
• I named the queries after their author in this thread, with an index where there are two.
• 在这个线程中，我以它们的作者命名查询，其中有一个索引，其中有两个。
• I ran all queries a couple of times to populate the cache, then I picked the best of 5 with EXPLAIN ANALYZE.
• 我运行所有查询几次以填充缓存，然后使用EXPLAIN ANALYZE选出其中最好的一个。

• Relevant indexes (should be the optimum - as long as we lack fore-knowledge which clubs will be queried):

  相关指标(应该是最优的——只要我们不知道哪些俱乐部会被查询):

  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 is 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:

  大多数查询都不需要club_pkey。主键在PostgreSQL中自动实现唯一索引。最后一个索引是为了弥补在PostgreSQL上的多列索引的已知缺点:

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.

一种多色的b -树索引可以与包含索引列的任何子集的查询条件一起使用，但是当对主要的(最左边的)列有约束时，索引是最有效的。

Results:

Total runtimes from EXPLAIN ANALYZE.

解释分析的总运行时间。

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.

最后三个表现差不多。4)和5)导致相同的查询计划。

Late Additions:

Fancy SQL, but the performance can't keep up.

花哨的SQL，但是性能跟不上。

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.

正如预期的那样，这两家公司的表现几乎相同。查询计划结果在表扫描中，计划者没有找到在这里使用索引的方法。

---

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 ...

漂亮的SQL，出色的CTE性能。非常奇异的查询计划。同样，有趣的是9.1如何处理这个。我将很快将这里使用的db集群升级到9.1。也许我会重新播放整部片子……

---

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.

令人惊讶的是，它可能会导致一个稍微不同的查询计划，并且具有完全相同的数据。我在student上发现了一个顺序扫描，其中的子查询变量使用了索引。

---

11) ypercube 3: 101.482 ms

Another late addition @ypercube. It is positively amazing, how many ways there are.

另一个加法@ypercube末。这很神奇，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
      )
   )

---

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.

@ypercube的11)实际上是这个更简单的变种的令人难以置信的反向方法，而这个变种至今仍未被发现。表现几乎和*猫一样快。

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
   )

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.

很难相信，但这是另一个真正的新变种。我认为可能会有两名以上的会员，但它也跻身只有两名会员的*猫科动物之列。

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
   )

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.

换句话说:改变过滤器的数量。这个问题正好要求两个俱乐部成员资格。但是许多用例必须为不同的数字做准备。

Detailed discussion in this related later answer:

在这个相关的后面的回答中有详细的讨论:

• Using same column multiple times in WHERE clause
• 在WHERE子句中多次使用同一列

#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

#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)

#4

If you just want student_id then:

如果您只想要student_id，则:

    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:

如果你也需要学生的名字，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 )

If you have more than two clubs in a club_selection table then:

如果在club_selection表中有两个以上的俱乐部，则:

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 ) )

#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)

或者更通用的解决方案，更容易扩展到n个俱乐部，避免交叉(MySQL中没有)和in(由于MySQL的性能很差)

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  

#6

Another CTE. It looks clean, but it will probably generate the same plan as a groupby in a normal subquery.

另一个CTE。它看起来很干净，但是它可能会在一个普通的子查询中生成与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)
    ;

For those who want to test, a copy of my generate testdata thingy:

对于那些想要测试的人，我的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)
    ;

#7

So there's more than one way to skin a cat.
I'll to add two more to make it, well, more complete.

所以剥猫皮的方法不止一种。我再加两个，让它更完整。

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:

假设一个健全的数据模型(student_id、club_id)在student_club中是唯一的。马丁·史密斯的第二版有点类似，但他先加入，之后再加入。这应该会更快:

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);

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):

当然，经典是存在的。类似于德里克的变种IN。简单和快速。(在MySQL中，这应该比带有In的版本快一些):

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);

#8

Since noone has added this (classic) version:

因为没有人加入这个(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  
              )
      )

or similar:

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  
              )
      )

---

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:

再试一试，用稍微不同的方法。受到解释扩展:EAV表中的多个属性的启发:GROUP by和NOT存在:

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:

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 ;   

#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.

这似乎表现得相当好，因为CTE-scan避免了两个单独的子查询。

There is always a reason to misuse recursive queries!

滥用递归查询总是有原因的!

(BTW: mysql does not seem to have recursive queries)

(顺便说一句:mysql似乎没有递归查询)

#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).

我在一个真实的数据库中测试过，所以名字和猫皮肤列表不同。它是一个备份副本，所以在所有测试运行期间没有任何更改(除了对目录的细微更改)。

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

#11

@erwin-brandstetter Please, benchmark this:

@erwin-brandstetter请基准:

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.

就像@sean写的6)一样，我猜他只是比较干净。

#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:

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)

So it still seems to want the seq scan on student.

所以它似乎仍然想要对学生进行seq扫描。

#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.

使用最快的变型(Mr. Sean in Mr. Brandstetter chart)。可能是变体，只有一个加入，只有学生俱乐部矩阵有生存的权利。所以，最长的查询只有两列来计算，想法是让查询变薄。