Question

我是MySQL新手;我正在使用SAS，我的大脑以SAS风格扭曲。所以，即使我认为我看到人们提出类似于我的问题，我仍然无法理解他们收到的答案。我为它道歉。

我有两张桌子：

包含列的film_category：

category_id,film_id,

，包含列的film_actor

actor_id, film_id.

我在film_id上加入了他们

。问题是：每个电影类别中最受欢迎的演员是什么？共有16个类别，200个演员，1000部电影。

SELECT film_actor.actor_id,MAX(COUNT(actor_id))
FROM film_actor JOIN film_category 
ON film_actor.film_id = film_category.film_id
GROUP BY film_category;

果然，只要我加上MAX，我就会收到错误1111 ... 如果我的问题不明确，请告诉我。谢谢！ -Vera

Answer 1

这是您的查询：

SELECT fa.actor_id, MAX(COUNT(fa.actor_id))
FROM film_actor fa JOIN
     film_category fc
     ON fa.film_id = fc.film_id
GROUP BY fc.film_category;

这是无效的，因为您无法嵌套聚合函数。

要做你想做的事，请使用子查询和聪明的聚合：

select film_category, max(numfilms) as maxnumfilms,
       substring_index(group_concat(actor_id order by numfilms desc), ',', 1) as actor_id
from (select fc.film_category, fa.actor_id, count(*) as numfilms
      from film_actor fa join
           film_category fc
           on fa.film_id = fc.film_id
      group by fc.film_category, fa.actor_id
     ) ac
group by film_category;

Answer 2

发生最多的演员： SELECT category_id AS cid，（SELECT actor_id FROM film_actor加入film_category ON film_actor.film_id = film_category.film_id WHERE category_id = cid GROUP BY actor_id 按COUNT（*）DESC排序 LIMIT 0,1）AS topActor 来自film_actor JOIN film_category ON film_actor.film_id = film_category.film_id GROUP BY category_id;

MySQL：MAX（COUNT（*）） - 错误1111

2 个答案: