Question

我遇到了MySQL查询的问题（我是新手，所以对我来说很容易！）。我有一个有电视节目的网站，想要根据两个不同的动作（自定义时间表）选择两个日期之间播出的剧集（没问题）。基本上，如果用户要么关注（uses_follow_shows）或观看节目（user_watched），我想要显示剧集，所以我想我会有类似下面的东西，但这不起作用。除了我的结果在一侧有一个查询的结果而在另一侧有另一个查询结果的事实之外，它的工作正常。

即：

    name                  a1    name            a
    Mistresses (US)        2    Wilfred (US)    2
    Mistresses (US)        2    Dexter          2

这是我的疑问：

我怎样才能使我的所有名字都在同一栏？

       SELECT * FROM (
         SELECT shows.name, users_follow_shows.user_id a1
       FROM show_episode_airdate join
            show_episode 
            on show_episode.episode_id = show_episode_airdate.episode_id join
            shows
            on shows.imdb_id = show_episode.imdb_id_show join
            show_network
            on show_network.show_id = shows.id join
            network
            on show_network.network_id = network.network_id join
            users_follow_shows
            on shows.id = users_follow_shows.show_id 

       WHERE show_episode_airdate.airdate BETWEEN '2013-07-20' AND '2013-07-27' and
             users_follow_shows.user_id = 2 

       ORDER by network.network_id ASC) a1

        JOIN (
         SELECT  shows.name, user_watched.user_id a
       FROM show_episode_airdate join
            show_episode 
            on show_episode.episode_id = show_episode_airdate.episode_id join
            shows
            on shows.imdb_id = show_episode.imdb_id_show join
            show_network
            on show_network.show_id = shows.id join
            network
            on show_network.network_id = network.network_id join
            user_watched
            on shows.id = user_watched.show_id 

       WHERE show_episode_airdate.airdate BETWEEN '2013-07-20' AND '2013-07-27' and
             user_watched.user_id = 2 

       ORDER by network.network_id ASC) a

       ON a = a1

任何帮助表示赞赏。感谢。

Answer 1

你不应该使用JOIN。您正在寻找的是UNION：

http://dev.mysql.com/doc/refman/5.0/en/union.html

Answer 2

你可以从这个子查询开始：

SELECT DISTINCT user_id, show_id
  FROM (
     SELECT user_id
            show_id
       FROM user_watched
     UNION
     SELECT user_id
            show_id
       FROM user_follows_show
  )

这将为您提供一个独特用户的虚拟表格，并显示符合您的两个条件，以便将它们包含在列表中。然后，您可以将其加入查询的其余部分，如下所示：

SELECT whatever
  FROM whatever
  JOIN (
    SELECT DISTINCT user_id, show_id
      FROM (
         SELECT user_id
                show_id
           FROM user_watched
         UNION
         SELECT user_id
                show_id
           FROM user_follows_show
      ) USER_INTERESTED ON USER_INTERESTED.show_id = shows.id
  WHERE whatever

诀窍是使用结构化查询语言的结构化部分来获取查询请求所需的用户/节目之间的虚拟关系。

Answer 3

未经测试但我会像这样简化您的查询：

SELECT 
    shows.name,
    users_follow_shows.user_id a1
FROM show_episode_airdate
    JOIN show_episode on show_episode.episode_id = show_episode_airdate.episode_id
    JOIN shows on shows.imdb_id = show_episode.imdb_id_show
    JOIN show_network on show_network.show_id = shows.id
    JOIN network on show_network.network_id = network.network_id
    LEFT JOIN users_follow_shows on shows.id = users_follow_shows.show_id
    LEFT JOIN user_watched on shows.id = user_watched.show_id
WHERE show_episode_airdate.airdate BETWEEN '2013-07-20' AND '2013-07-27'
    AND (users_follow_shows.user_id = 2 OR user_watched.user_id = 2)

结合两个查询OR和LEFT Join - MySQL

3 个答案: