Question

我需要选择大多数电影中出现的演员。我正在使用sakila数据库1.0：

.mwb (MySQL Workbench): https://www.sendspace.com/file/i0z82j
.sql (schema): https://www.sendspace.com/file/vd3hnu
.sql (data): https://www.sendspace.com/file/gbp9ri

我已经拥有的是两个查询

选择每个演员n次（n =电影中的出场次数）。我的查询： SELECT actor.first_name，actor.last_name 来自演员 INNER JOIN film_actor ON actor.actor_id = film_actor.actor_id;

结果：

+------------+-------------+
| first_name | last_name   |
+------------+-------------+
| PENELOPE   |GUINESS      | //4 Films in this examle
| PENELOPE   |GUINESS      |
| PENELOPE   |GUINESS      |
| PENELOPE   |GUINESS      |
| NICK       |WAHLBERG     | //5 Films in this examle
| NICK       |WAHLBERG     |
| NICK       |WAHLBERG     |
| NICK       |WAHLBERG     |
| NICK       |WAHLBERG     |
| ED         |CHASE        | //5 Films in this examle
| ED         |CHASE        |
| ED         |CHASE        |
| ED         |CHASE        |
| ED         |CHASE        |
| JENNIFER   |DAVIS        | //4 Films in this examle
| JENNIFER   |DAVIS        |
| JENNIFER   |DAVIS        |
| JENNIFER   |DAVIS        |
| BETTE      |NICHOLSON    | //3 Films in this examle
| BETTE      |NICHOLSON    |
| BETTE      |NICHOLSON    |
| ...        |...          | //Everyother Actor who appeared in a film
+------------+-------------+

选择特定演员出现的每部电影。我的查询： SELECT actor_id，film_id 来自film_actor WHERE actor_id = 1;

结果：演员1出现在19部电影中。

+----------+-------- +
| actor_id | film_id |
+----------+---------+
| 1        | 1       |
| 1        | 23      |
| 1        | 25      |
| 1        | 106     |
| 1        | 140     |
| 1        | 166     |
| 1        | 277     |
| 1        | 361     |
| 1        | 438     |
| 1        | 499     |
| 1        | 506     |
| 1        | 509     |
| 1        | 605     |
| 1        | 635     |
| 1        | 749     |
| 1        | 832     |
| 1        | 939     |
| 1        | 970     |
| 1        | 980     |
+----------+---------+

但我想要的是：出现在大多数电影中的演员：

+------------+---------------+--------+
| first_name | last_name     | films  |
+------------+---------------+--------+
| NICK       | WAHLBERG      | 5      |
| ED         | CHASE         | 5      |
| ...        | ...           | ...    |
+------------+----------------+-------+

非常感谢任何帮助。谢谢！

Answer 1

无法在您的数据库上运行查询我会这样做：

SELECT count(film_actor.actor_id), actor.first_name, actor.last_name 
FROM actor INNER JOIN film_actor ON actor.actor_id = film_actor.actor_id
GROUP BY film_actor.actor_id ;

未测试

Answer 2

select concat(a.FIRST_NAME,' ', a.LAST_NAME) as Full_name
from ACTOR a
left join FILM_ACTOR fa on a.ACTOR_ID = fa.ACTOR_ID
group by a.ACTOR_ID, a.FIRST_NAME, a.LAST_NAME
order by count(*) desc
limit 1;

MySQL Sakila选择最多电影的演员

2 个答案: