Question

使用：http://dev.mysql.com/doc/sakila/en/

我试图写一个查询来计算每个演员按类别行事的电影数量。

非常感谢任何指示或建议。

到目前为止，这是我的代码（当我拿出第二个LEFT JOIN部分时，它正确计算了一个类别的影片）：

SELECT actor.first_name, actor.last_name, 
COUNT(subset.film_id) AS action,
COUNT(subset2.film_id) AS animation
FROM actor
LEFT JOIN (
SELECT film.film_id, actor.actor_id 
FROM actor
INNER JOIN film_actor
ON film_actor.actor_id = actor.actor_id
INNER JOIN film 
ON film.film_id = film_actor.film_id
INNER JOIN film_category 
ON film_category.film_id = film.film_id
INNER JOIN category 
ON category.category_id = film_category.category_id
WHERE category.name = 'Action') AS subset 
ON subset.actor_id = actor.actor_id
LEFT JOIN (
SELECT film.film_id, actor.actor_id 
FROM actor
INNER JOIN film_actor
ON film_actor.actor_id = actor.actor_id
INNER JOIN film 
ON film.film_id = film_actor.film_id
INNER JOIN film_category 
ON film_category.film_id = film.film_id
INNER JOIN category 
ON category.category_id = film_category.category_id
WHERE category.name = 'Animation') AS subset2
ON subset2.actor_id = actor.actor_id
GROUP BY actor.actor_id
ORDER BY actor.last_name ASC;

Answer 1

如果您只想获得每个演员和类别的电影数量，那么您所获得的查询似乎非常复杂。

此查询：

SELECT 
    actor.first_name, actor.last_name, category.name, 
    COUNT(*) as CountPerCategory
FROM actor
JOIN film_actor ON film_actor.actor_id = actor.actor_id
JOIN film ON film.film_id = film_actor.film_id
JOIN film_category ON film_category.film_id = film.film_id
JOIN category ON category.category_id = film_category.category_id
GROUP BY actor.first_name, actor.last_name, category.name;

会给你一个输出：

Firstname, Lastname, Category, CountPerCategory
Clint      Eastwood  Animation 1
Clint      Eastwood  Action    15

但是如果你想要不同列中每个类别的计数，你可以利用MySQL为真实条件返回1的事实，并测试该类别是否匹配并使用sum（这也可能不包括案例表达式）一种更便携的方式），像这样：

SELECT 
    actor.first_name, actor.last_name, category.name
    , SUM(category.name = 'Animation') as CountOfAnimation
    , SUM(category.name = 'Action') as CountOfAction
FROM actor
JOIN film_actor ON film_actor.actor_id = actor.actor_id
JOIN film ON film.film_id = film_actor.film_id
JOIN film_category ON film_category.film_id = film.film_id
JOIN category ON category.category_id = film_category.category_id
GROUP BY actor.first_name, actor.last_name, category.name;

会产生如下结果：

Firstname, Lastname, CountOfAnimation, CountOfAction
Clint      Eastwood  1                 15

请注意，如果一部电影属于多个类别，则每个类别都会计算一次，这可能是您想要的（或不是）。

Answer 2

这应该为您提供所有参与者的计数，包括与之关联的所有类别：

SELECT actor.first_name, actor.last_name, category.name
, COUNT(DISTINCT film.film_id)
FROM actor
LEFT JOIN film_actor ON film_actor.actor_id = actor.actor_id
LEFT JOIN film ON film.film_id = film_actor.film_id
LEFT JOIN film_category ON film_category.film_id = film.film_id
LEFT JOIN category ON category.category_id = film_category.category_id
GROUP BY actor.first_name, actor.last_name, category.name
;

如果您只想要特定的，请在GROUP BY行之后添加：

HAVING category.name IN ('Action', 'Animation')

如果您只想要拥有这些类别的演员，请将LEFT JOIN更改为INNER JOIN和/或将上述HAVING更改为WHERE（并相应地重新定位）。

Answer 3

我同意，该查询看起来非常复杂，并且通过使用2个派生表，它也可能运行缓慢。此查询将在每个类别中提供唯一的电影，并且不会复制结果，因为没有多对多的连接。它会在您添加类别时减慢，但是因为它必须多次扫描表格，但它会给出准确的结果，只要您的表格不是太大就应该没问题。

你可能不需要不同的，但如果出于某种原因，某人在电影上有两次相同的类别（或类似的东西），这将消除错误的空间。

select a.first_name, a.last_name
, (
select count(distinct f.id) 
from film f 
join film_actor fa on fa.film_id=f.film_id
join film_category fc on fc.film_id=f.film_id
join category c ON c.category_id = fc.category_id
where c.name='Animation'
and fa.actor_id=a.actor_id
) as unique_animation_films
, (
select count(distinct f.id) 
from film f 
join film_actor fa on fa.film_id=f.film_id
join film_category fc on fc.film_id=f.film_id
join category c ON c.category_id = fc.category_id
where c.name='Action'
and fa.actor_id=a.actor_id
) as unique_action_films
from actor a

具有多个左连接的SQL

3 个答案: