考虑以下几点：
DROP TABLE IF EXISTS person;
CREATE TABLE person
(person_id SERIAL PRIMARY KEY
,name VARCHAR(20) NOT NULL UNIQUE
);
DROP TABLE IF EXISTS movie;
CREATE TABLE movie
(movie_id SERIAL PRIMARY KEY
,title VARCHAR(50) NOT NULL UNIQUE
);
DROP TABLE IF EXISTS m_cast;
CREATE TABLE m_cast
(movie_id INT NOT NULL
,person_id INT NOT NULL
,PRIMARY KEY(movie_id,person_id)
);
DROP TABLE IF EXISTS m_director;
CREATE TABLE m_director
(movie_id INT NOT NULL
,person_id INT NOT NULL
,PRIMARY KEY(movie_id,person_id)
);
INSERT INTO person (name) VALUES
('Steven Feelberg'),
('Manly Kubrick'),
('Alfred Spatchcock'),
('Fred Pitt'),
('Raphael DiMaggio'),
('Bill Smith');
INSERT INTO movie VALUES
(1,'Feelberg\'s Movie with Fred & Raph'),
(2,'Feelberg and Fred Ride Again'),
(3,'Kubrick shoots DiMaggio'),
(4,'Kubrick\'s Movie with Bill Smith'),
(5,'Spatchcock Presents Bill Smith');
INSERT INTO m_director VALUES
(1,1),
(2,1),
(3,2),
(4,2),
(5,3);
INSERT INTO m_cast VALUES
(1,4),
(1,5),
(2,4),
(3,5),
(4,6),
(5,6);
我包含电影表只是为了方便参考。它与实际问题无关。
另请注意，此模型假定演员仅列出一次，无论他们是否在给定电影中担任多个角色。
以下查询询问“每个演员和导演多久合作一次”...
演员是任何电影的演员。
导演是任何电影导演的人。
SELECT a.name actor
, d.name director
, COUNT(DISTINCT ma.movie_id) total
FROM person d
JOIN m_director md
ON md.person_id = d.person_id
JOIN person a
LEFT
JOIN m_cast ma
ON ma.person_id = a.person_id
AND ma.movie_id = md.movie_id
JOIN m_cast x
ON x.person_id = a.person_id
GROUP
BY actor
, director;
+-------------------+-------------------+-------+
| actor | director | total |
+-------------------+-------------------+-------+
| Fred Pitt | Alfred Spatchcock | 0 |
| Fred Pitt | Manly Kubrick | 0 |
| Fred Pitt | Steven Feelberg | 2 |
| Raphael DiMaggio | Alfred Spatchcock | 0 |
| Raphael DiMaggio | Manly Kubrick | 1 |
| Raphael DiMaggio | Steven Feelberg | 1 |
| Bill Smith | Alfred Spatchcock | 1 |
| Bill Smith | Manly Kubrick | 1 |
| Bill Smith | Steven Feelberg | 0 |
+-------------------+-------------------+-------+
通过观察，我们可以看到：
- 唯一一位比其他导演更常与菲尔伯格合作的演员是弗雷德·普里特
- 拉斐尔·迪卡普里奥和比尔·史密斯都与两位导演（尽管是不同的导演）合作频繁
编辑：虽然我并没有认真提倡将此作为解决方案，但以下只是为了证明上面提供的内核确实是您解决问题所需要的一切......
SELECT x.*
FROM
( SELECT a.*
FROM
( SELECT a.name actor
, d.name director
, COUNT(DISTINCT ma.movie_id) total
FROM person d
JOIN m_director md
ON md.person_id = d.person_id
JOIN person a
LEFT
JOIN m_cast ma
ON ma.person_id = a.person_id
AND ma.movie_id = md.movie_id
JOIN m_cast x
ON x.person_id = a.person_id
GROUP
BY actor
, director
) a
LEFT
JOIN
( SELECT a.name actor
, d.name director
, COUNT(DISTINCT ma.movie_id) total
FROM person d
JOIN m_director md
ON md.person_id = d.person_id
JOIN person a
LEFT
JOIN m_cast ma
ON ma.person_id = a.person_id
AND ma.movie_id = md.movie_id
JOIN m_cast x
ON x.person_id = a.person_id
GROUP
BY actor
, director
) b
ON b.actor = a.actor
AND b.director <> a.director
AND b.total > a.total
WHERE b.actor IS NULL
) x
LEFT JOIN
( SELECT a.*
FROM
( SELECT a.name actor
, d.name director
, COUNT(DISTINCT ma.movie_id) total
FROM person d
JOIN m_director md
ON md.person_id = d.person_id
JOIN person a
LEFT
JOIN m_cast ma
ON ma.person_id = a.person_id
AND ma.movie_id = md.movie_id
JOIN m_cast x
ON x.person_id = a.person_id
GROUP
BY actor
, director
) a
LEFT
JOIN
( SELECT a.name actor
, d.name director
, COUNT(DISTINCT ma.movie_id) total
FROM person d
JOIN m_director md
ON md.person_id = d.person_id
JOIN person a
LEFT
JOIN m_cast ma
ON ma.person_id = a.person_id
AND ma.movie_id = md.movie_id
JOIN m_cast x
ON x.person_id = a.person_id
GROUP
BY actor
, director
) b
ON b.actor = a.actor
AND b.director <> a.director
AND b.total > a.total
WHERE b.actor IS NULL
) y
ON y.actor = x.actor AND y.director <> x.director
WHERE y.actor IS NULL;
+-----------+-----------------+-------+
| actor | director | total |
+-----------+-----------------+-------+
| Fred Pitt | Steven Feelberg | 2 |
+-----------+-----------------+-------+
这会返回每个演员的列表，以及他们最常合作的导演。在这种情况下，由于 Bill Smith 和 Raphael DiMaggio 与两位导演合作的次数最多，因此他们被排除在结果之外。
您的问题的答案是简单地从该列表中选择 Yash Chopra 列为主管的所有行。