Question

我发现了类似的问题，但最后却完全不同，没有人回答我的问题。

我使用mySQL

我有一张桌子movie_artist_role喜欢3桌电影，艺术家和包含3把钥匙的角色

mo_id
ar_id
ro_id.

表格电影包含

mo_id
mo_title

表格艺术家包含

ar_id
ar_name

表角色包含

ro_id
ro_name (actor or producer for instance)

我想要显示电影列表，以及演员和导演的每部电影

最有效的方法是什么？是否可以在一个具有多个左连接的唯一查询中？

由于

Answer 1

select movie.mo_title, artist.ar_name, role.ro_name
from movie_artist_role 
left join movie on movie.mo_id = movie_artist_role.mo_id
left join artist on artist.ar_id = movie_artist_role.ar_id
left join role on role.ro_id = movie_artist_role.ro_id
order by movie.mo_tile, artist.ar_name

我有一个类似的电影数据库！我认为你所定义的'角色'表是没有必要的 - 我有一个单独的'导演'表直接链接到'电影'表：很少有两个导演用于同一个moview，当有（例如科恩兄弟），他们往往经常一起工作，所以他们可以被视为一个导演。

使用左连接而不是内部连接可以显示没有导演或连接到它的演员的电影。

Answer 2

连接的inner join之间的多个foreign keys（将从所有3个表中提取所有内容）将显示电影，演员和导演的信息列表。

SELECT * FROM movie_artist_role mar
INNER JOIN movie m ON mar.mo_id = m.mo_id
INNER JOIN artists a ON mar.ar_id = a.ar_id
INNER JOIN role r ON mar.ro_id = r.ro_id
ORDER BY m.mo_title, r.role_name

使用WHERE子句可以为您提供有关该电影或演员的详细信息。例如..

SELECT * FROM movie_artist_role mar
INNER JOIN movie m ON mar.mo_id = m.mo_id
INNER JOIN artists a ON mar.ar_id = a.ar_id
INNER JOIN role r ON mar.ro_id = r.ro_id
WHERE a.ar_name = 'Jack Black' ORDER BY m.mo_title, r.role_name

如果您只想要来自特定表格的特定数据，请说film然后......它几乎是上面所做的，但反之亦然。

SELECT * FROM movie m
INNER JOIN movie_artist_role mar ON mar.mo_id = m.mo_id
INNER JOIN artists a ON mar.ar_id = a.ar_id
INNER JOIN role r ON mar.ro_id = r.ro_id
ORDER BY m.mo_title, r.role_name

或artists ...

SELECT * FROM artists a
INNER JOIN movie_artist_role mar ON mar.ar_id = a.ar_id
INNER JOIN movie m ON mar.mo_id = m.mo_id
INNER JOIN role r ON mar.ro_id = r.ro_id
ORDER BY m.mo_title, r.role_name

等等

Answer 3

LEFT JOINS应该可以正常工作。

SELECT m.mo_title AS movie_title, 
       (SELECT group_concat(aIn.ar_name)
        FROM movie_artist_role AS marIn
            LEFT JOIN role AS rIn ON rIn.ro_id = marIn.ro_id
            LEFT JOIN artist AS aIn ON aIn.ar_id = marIn.ao_id
        WHERE rIn.ro_name = 'producer'
            AND marIn.mo_id = m.id ) AS producers,

       (SELECT group_concat(aIn.ar_name)
        FROM movie_artist_role AS marIn
            LEFT JOIN role AS rIn ON rIn.ro_id = marIn.ro_id
            LEFT JOIN artist AS aIn ON aIn.ar_id = marIn.ao_id
        WHERE rIn.ro_name = 'actor'
            AND marIn.mo_id = m.id ) AS actors
FROM movie AS m

这是未经测试的，但应返回如下行：

movie_title       |      producers       |     actors
-----------------------------------------------------------
Star Treck          JJ Abrams               Chris Pine, Simon Pegg, Zachery

SQL - 从电影数据库请求获取所有电影及其详细信息

3 个答案: