问题描述
在IMBD数据库中查找使用Yash Chopra制作的电影多于其他任何导演的所有演员。
示例架构：
person
(pid *,name
);
m_cast
(mid *,pid *
);
m_director
(mid*,pid*
);
* = (component of) PRIMARY KEY
以下是我的查询：
WITH common_actors AS
(SELECT A.actor_id as actors,B.director_id as director_id,B.movies as movies_with_director,B.director_id as yash_chops_id,B.movies as movies_with_yash_chops FROM
(SELECT M_Cast.PID as actor_id,M_Director.PID as director_id,COUNT(*) as movies from M_Cast
left join M_Director
ON M_Cast.MID = M_Director.MID
GROUP BY actor_id,director_id) A
JOIN
(SELECT M_Cast.PID as actor_id,director_id
)B
ON A.actor_id = B.actor_id
WHERE B.director_id in (SELECT PID FROM Person WHERE Name LIKE
'%Yash%Chopra%'))
SELECT distinct actors as actor_id,movies_with_yash_chops as total_movies FROM common_actors
WHERE actors NOT IN (SELECT actors FROM common_actors WHERE movies_with_director > movies_with_yash_chops)
从中获得的结果的长度为：430行。但是，获得的结果应为243行。谁能建议我在查询中出了错吗？我的方法正确吗？
抽样结果：
Actor name
0 Sharib Hashmi
1 Kulbir Badesron
2 Gurdas Maan
3 Parikshat Sahni
...
242 Ramlal Shyamlal
谢谢！
解决方法
请考虑以下内容：
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)
);
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),3);
INSERT INTO m_cast VALUES
(1,4),(1,5),6),6);
我包括电影表只是为了便于参考。与实际问题无关。 另外，请注意，此模型假定演员仅列出一次，无论他们在给定电影中是否具有多个角色。
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。 导演是担任过任何电影导演的任何人。
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：
- 弗雷德·普里特（Fred Pritt）是与费尔伯格一起工作次数最多的唯一演员
- 拉斐尔·迪卡普里奥（Raphael DiCaprio）和比尔·史密斯（Bill Smith）经常与两名董事（尽管董事不同）共同工作
编辑：虽然我没有认真地倡导将其作为解决方案，但以下内容只是为了证明上述提供的内核确实是解决问题所需要的...
SELECT x.*
FROM
( SELECT a.*
FROM
( SELECT a.name actor,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,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,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 |
+-----------+-----------------+-------+
这将返回每个演员的名单，以及与他们最常合作的导演。在这种情况下，由于比尔·史密斯和拉斐尔·迪马吉奥最经常与两位董事平等地合作，因此他们被排除在结果之外。
您的问题的答案只是从列表中选择列出Yash Chopra作为主管的所有行。