Question

我想找到每种类型的电影，找到在大多数类型的电影中播放的N个演员

Tables and their columns:

actor(actor_id,name)
role(actor_id,movie_id)
movie(movie_id,title)
movie_has_genre(movie_id,genre_id)
genre(genre_id,genre_name)

通过此查询，我可以找到播放同一类型电影的演员。

select t1.genre_name, t1.actor_id, t1.max_value
from
(
    select g.genre_name, a.actor_id, count(*) as max_value
    from genre g
    inner join movie_has_genre mhg on mhg.genre_id = g.genre_id
    inner join movie m on mhg.movie_id = m.movie_id
    inner join role r on m.movie_id = r.movie_id
    inner join actor a on a.actor_id = r.actor_id
    group by g.genre_name, a.actor_id
) t1
inner join
(
    select genre_name, MAX(max_value) AS max_value
    from
    (
        select g.genre_name, a.actor_id, count(*) as max_value
        from genre g
        inner join movie_has_genre mhg on mhg.genre_id = g.genre_id
        inner join movie m on mhg.movie_id = m.movie_id
        inner join role r on m.movie_id = r.movie_id
        inner join actor a on a.actor_id = r.actor_id
        group by g.genre_name, a.actor_id
    ) t
    GROUP BY genre_name
) t2
    ON t1.genre_name = t2.genre_name and t1.max_value = t2.max_value
ORDER BY
    t1.max_value desc;

但是我想把演员的数量限制为1.那我怎么能这样做呢？

示例：

我得到的结果：

genre_name | actor_id | max_value
==================================
Thriller   |  22591   |   7
Drama      |  22591   |   6
Crime      |  65536   |   3
Horror     |  22591   |   3
Action     |  292028  |   3
Action     |  378578  |   3
Action     |  388698  |   3

我想要的结果：

 genre_name | actor_id | max_value
==================================
Thriller    | 22591    | 7
Drama       | 22591    | 6
Crime       | 65536    | 3
Horror      | 22591    | 3
Action      | 292028   | 3

Answer 1

如果您只想随机选择一个演员，只需在代码中添加以下行：

 select genre_name, actor_id, max_value
 from
 (
     select g.genre_name, a.actor_id, count(*) as max_value
     from genre g
     inner join movie_has_genre mhg on mhg.genre_id = g.genre_id
     inner join movie m on mhg.movie_id = m.movie_id
     inner join role r on m.movie_id = r.movie_id
     inner join actor a on a.actor_id = r.actor_id
     group by g.genre_name, a.actor_id
 ) t1
 inner join
 (
     select genre_name, MAX(max_value) AS max_value
     from
     (
         select g.genre_name, a.actor_id, count(*) as max_value
         from genre g
         inner join movie_has_genre mhg on mhg.genre_id = g.genre_id
         inner join movie m on mhg.movie_id = m.movie_id
         inner join role r on m.movie_id = r.movie_id
         inner join actor a on a.actor_id = r.actor_id
         group by g.genre_name, a.actor_id
     ) t
     GROUP BY genre_name
 ) t2
     USING(genre_name,max_value)
 GROUP BY genre_name, max_value
 ORDER BY max_value desc;

Answer 2

您使用的某些联接是多余的。

SELECT 
  U.genre_name, U.actor_id, U.actor_genre_count
FROM
    (SELECT 
      A.genre_id, A.genre_name, C.actor_id, count(*) actor_genre_count
    FROM genre A 
    JOIN movie_has_genre B
    ON A.genre_id=B.genre_id 
    JOIN role C 
    ON C.movie_id=B.movie_id
    GROUP BY A.genre_id, A.genre_name, C.actor_id) U
JOIN 
   (SELECT 
       S.genre_id, S.genre_name, MAX(S.actor_genre_count) max_actor_genre
    FROM
       (SELECT 
           A.genre_id, A.genre_name, C.actor_id, count(*) actor_genre_count
        FROM genre A
        JOIN movie_has_genre B
        ON A.genre_id=B.genre_id 
        JOIN role C 
        ON C.movie_id=B.movie_id
        GROUP BY A.genre_id, A.genre_name, C.actor_id) S
    GROUP BY S.genre_id, S.genre_name) V
ON U.genre_name=V.genre_name AND U.actor_genre_count=V.max_actor_genre;

Answer 3

该解决方案改编自this Stack Overflow answer关于限制名称的结果。我试图做一个类似的查询，应该选择第一个actor_id并且只返回它。

SELECT id, CategoryName, image, date_listed, item_id
SELECT t1.genre_name, t1.actor_id, t1.actor_movie_count
FROM
    (
    SELECT g.genre_name, r.actor_id, COUNT(*) as actor_movie_count    
    FROM genre g
    INNER JOIN movie_has_genre mhg ON mhg.genre_id = g.genre_id
    INNER JOIN role r ON m.movie_id = r.movie_id
    GROUP BY g.genre_name, r.actor_id
    ) t1
LEFT JOIN
    (
    SELECT genre_name, actor_id, MAX(actor_movie_count) AS max_actor_movie_count
    FROM
         (
         SELECT g.genre_name, r.actor_id, COUNT(*) AS actor_movie_count
         FROM genre g
         INNER JOIN movie_has_genre mhg ON mhg.genre_id = g.genre_id
         INNER JOIN role r ON m.movie_id = r.movie_id
         GROUP BY g.genre_name, r.actor_id
         )
    GROUP BY genre_name
    ) t2
ON t1.genre_name = t2.genre_name AND t1.actor_movie_count = t2.max_actor_movie_count AND (t1.actor_id > t2.actor_id)
WHERE t2.genre_id IS NULL
ORDER BY t1.actor_movie_count DESC

如果这仍然无法解决您的问题，其他类似的问题及解释说明如下：

SO answer about returning 1 row per group

SO question about limiting query answer to N results per group

SO question about selecting N items per category

外部文章：Finding the max/first of a particular group in SQL

Answer 4

您可以使用相关的LIMIT 1子查询来获取播放该类型的演员的id。

select g.genre_name, (
        select r.actor_id
        from movie_has_genre mg
        join role r on r.movie_id = mg.movie_id
        where mg.genre_id = g.genre_id
        group by r.actor_id
        order by count(*) desc,
                 r.actor_id asc -- on tie least actor_id wins
    ) as actor_id
from genre g

结果如下：

genre_name | actor_id
======================
Thriller   | 22591   
Drama      | 22591   
Crime      | 65536   
Horror     | 22591   
Action     | 292028

如您所见，计数不包括在内。如果您需要计数，那么简单的方法是使用actor_id

将其返回到相同的字符串列中

将子查询中的SELECT子句更改为

select concat(r.actor_id, ':', count(*)) as actor_id_count

这将返回actor_id和单个字符串列中的计数，如

genre_name | actor_id_count
===========================
Thriller   |  22591:7

然后，您可以在应用程序代码中解析它（使用split，explode或其他任何内容。

CTE （公用表表达式）和ROW_NUMBER()（窗口函数）的解决方案（由 MySQL支持） 8 和 MariaDB 10.2 ）可能是：

with cte as (
    select g.genre_name, r.actor_id, count(*) as max_value,
       row_number() over (partition by g.genre_name order by count(*) desc, r.actor_id) as rn
    from genre g
    inner join movie_has_genre mhg on mhg.genre_id = g.genre_id
    inner join role r on mhg.movie_id = r.movie_id
    group by g.genre_name, r.actor_id
)
select genre_name, actor_id, max_value from cte where rn = 1

如何限制最大数量

4 个答案: