MySQL：多个子查询的总和答案

【问题标题】：MySQL: Sum of multiple subqueriesMySQL：多个子查询的总和
【发布时间】：2014-04-12 04:22:49
【问题描述】：

我需要通过多个子查询的总和来订购MySQL 查询。

这是我正在尝试做的一些示例代码：

SELECT  ...
        (SELECT SUM(
           (SELECT one_result ... LIMIT 1) as plays1,
           (SELECT one_result ... LIMIT 1) as plays2,
           (SELECT one_result ... LIMIT 1) as plays3
        )) as total_plays
FROM plays
ORDER BY total_plays

基本上，我需要运行三个子查询，每个子查询都会返回一个整数。

我需要通过这些整数的SUM() 对整个查询进行排序。

当我尝试运行此查询时，出现语法错误。

谁能告诉我对多个子查询求和的正确语法是什么？

我也试过了：

SELECT  ...
        (SELECT one_result ... LIMIT 1) as plays1,
        (SELECT one_result ... LIMIT 1) as plays2,
        (SELECT one_result ... LIMIT 1) as plays3
        SUM(plays1, plays3, plays3) as total_plays
FROM plays
ORDER BY total_plays

编辑

@JoeC 和 @SATSON 提供了解决此问题的类似答案。这是我的（工作）真实查询，以防其他人开始他们自己的查询：

````

SELECT  song.title as title,
        song.file_name as unique_name,
        song.artist as artist,
       (SELECT difficulty FROM chart WHERE id = song.easy_chart ORDER BY scoring_version ASC LIMIT 1) as easy_difficulty,
       (SELECT difficulty FROM chart WHERE id = song.hard_chart ORDER BY scoring_version ASC LIMIT 1) as hard_difficulty,
       (SELECT difficulty FROM chart WHERE id = song.master_chart ORDER BY scoring_version ASC LIMIT 1) as master_difficulty,
       (plays.easy_plays + plays.hard_plays + plays.master_plays) as total_plays
FROM song
INNER JOIN (SELECT parent_song_id,
               (SELECT global_plays ORDER BY scoring_version ASC LIMIT 1) as easy_plays,
               (SELECT global_plays ORDER BY scoring_version ASC LIMIT 1) as hard_plays,
               (SELECT global_plays ORDER BY scoring_version ASC LIMIT 1) as master_plays
       FROM chart) as plays
ON plays.parent_song_id = song.id
ORDER BY total_plays DESC
LIMIT 9
OFFSET 0

````

【问题讨论】：

标签： mysql sum subquery

【解决方案1】：

嗯，怎么样

SELECT *, plays1 + plays2 + plays3 as total_play from 
(SELECT  ...
        (SELECT one_result ... LIMIT 1) as plays1,
        (SELECT one_result ... LIMIT 1) as plays2,
        (SELECT one_result ... LIMIT 1) as plays3
FROM plays)
ORDER BY total_plays

【讨论】：

#1054 - “字段列表”中的未知列“plays1”
那行得通。对于任何感兴趣的人，我已经编辑了我的问题以显示真实的查询，以防其他人形成他们自己的查询。

【解决方案2】：

这样试试

SELECT  sum(plays1) as total_plays from (
        (SELECT one_result as plays1  ... LIMIT 1)
        union all 
        (SELECT one_result as plays1 ... LIMIT 1 )
        union all
        (SELECT one_result as plays1 ... LIMIT 1)
      )
as  plays
ORDER BY total_plays

【讨论】：