Mysql 使用复杂子查询在SQL中练习

我正在尝试用SQL进行查询。该数据库是IMBD数据库中的一个。表格如下：

所以，我需要电影《低俗小说》中的男演员/女演员ID=2175869，他们从未在其他电影中与另一个低俗小说的男演员/女演员合作

应该是这样的：

SELECT person_id FROM cast_info WHERE movie_id = 2175869 AND
person_id NOT IN (SELECT )

---

但是我不知道如何做第二部分。

查询稍后用作子查询，返回除低俗小说以外的电影，其中有一个以上的低俗小说演员在其中扮演

SELECT movie_id
FROM cast_info
WHERE
    movie_id <> 2175869 AND
    person_id IN (SELECT person_id FROM cast_info WHERE movie_id = 2175869)
GROUP BY movie_id
HAVING COUNT(*) > 1

现在让那些从未在这样一部电影中出演过低俗小说的人来看看

SELECT person_id
FROM cast_info
WHERE
    movie_id = 2175869 AND
    person_id NOT IN (
        SELECT person_id FROM cast_info WHERE movie_id IN (
            SELECT movie_id
            FROM cast_info
            WHERE
                movie_id <> 2175869 AND
                person_id IN (SELECT person_id FROM cast_info WHERE movie_id = 2175869)
            GROUP BY movie_id
            HAVING COUNT(*) > 1
        )
    )

---

注：我假设每部电影只能选一名演员。如果同一个演员可以在同一部电影中扮演不同的角色，则用COUNTDISTINCT person_id替换COUNT*。

首先，您需要查看《低俗小说》中的所有演员：

SELECT person_id  as pulp_actor_id
FROM cast_info
WHERE movie_id = 2175869

然后你需要找到所有演员对决：

SELECT c1.person_id as actor1_id
       c2.person_id as actor2_id
FROM cast_info c1
JOIN cast_info c2
  ON c1.person_id < c2.person_id
WHERE c1.movie_id = 2175869
  AND c2.movie_id = 2175869;

找到除了低俗小说之外，两人一起表演的所有电影

SELECT actor1_id, actor2_id, actor1_movies.movie_id
FROM (SELECT c1.person_id as actor1_id
             c2.person_id as actor2_id
      FROM cast_info c1
      JOIN cast_info c2
        ON c1.person_id < c2.person_id
      WHERE c1.movie_id = 2175869
        AND c2.movie_id = 2175869) as duos
JOIN cast_info as actor1_movies
  ON duos.actor1_id = actor1_movies.person_id
 AND actor1.movie_id <> 2175869
LEFT JOIN cast_info actor2_movies
  ON duos.actor2_id = actor2_movies.person_id
 AND actor1_movies.movie_id = actor2_movies.movie_id
WHERE actor2.movie_id IS NOT NULL;

现在把所有的演员都从低俗小说中选出来，而不是最后的结果

SELECT person_id as pulp_actor_id
FROM cast_info c
LEFT JOIN ( SELECT actor1_id, actor2_id, actor1_movies.movie_id
            FROM (SELECT c1.person_id as actor1_id
                         c2.person_id as actor2_id
                  FROM cast_info c1
                  JOIN cast_info c2
                    ON c1.person_id < c2.person_id
                  WHERE c1.movie_id = 2175869
                    AND c2.movie_id = 2175869) as duos
            JOIN cast_info as actor1_movies
              ON duos.actor1_id = actor1_movies.person_id
             AND actor1.movie_id <> 2175869
            LEFT JOIN cast_info actor2_movies
              ON duos.actor2_id = actor2_movies.person_id
             AND actor1_movies.movie_id = actor2_movies.movie_id
            WHERE actor2.movie_id IS NOT NULL;
            WHERE movie_id = 2175869 ) as duos
  ON c.person_id IN ( actor1_id, actor2_id )
WHERE duos.movie_id IS NULL

---

@萨米：我不太确定。你指的是与@Harshil相同的问题吗？是的，他指的是与mineOkey相同的问题。好的，谢谢，我现在正在学习。我找不到一张表格，上面列出了一个人/导演/演员的所有电影电影id。请帮帮我。@JaimeAlcántaraArnela你将如何匹配这个title.id=cast\u info.person\u id？这两个都是不同的na？谢谢，我会试试，但这需要时间，我会把结果发布给你。Hola Jaime，检查Olver的答案，比我的简单。Muy honesto por tu parte Juan Carlos，gracias，voy a elloChecking，这是一个非常大的数据库，所以需要时间，但它运行正常。知道，我迟早会发布结果的。谢谢