Question

这是我正在尝试使用的查询的精简版本：

SELECT user_id,
(SELECT COUNT(*) FROM responses WHERE option_id = 1 AND poll_id = 1 AND pref = 1) /
(SELECT COUNT(*) FROM responses WHERE poll_id = 1 AND pref = 1) AS pref_percent,
(SELECT COUNT(*) FROM responses WHERE option_id = 1 AND poll_id = 1 ) /
(SELECT COUNT(*) FROM responses WHERE poll_id = 1) AS all_percent,
ABS(all_percent - pref_percent) AS deviation
FROM responses
GROUP BY user_id
ORDER BY deviation DESC

我收到此错误：'字段列表'中的未知列'all_percent'

显然，在构造另一个用户定义的列（“偏差”）时，我无法引用用户定义的列（“all_percent”）。

所以，我的问题是......我如何修改此查询以避免错误？

我真的希望将查询结果按偏差排序，而不是必须在PHP中完成这项工作。

Answer 1

围绕当前查询包装外部查询：

SELECT user_id, pref_percent, all_percent, ABS(all_percent - pref_percent) AS deviation
    FROM (SELECT user_id,
          (SELECT COUNT(*) FROM responses WHERE option_id = 1 AND poll_id = 1 AND pref = 1) /
          (SELECT COUNT(*) FROM responses WHERE poll_id = 1 AND pref = 1) AS pref_percent,
          (SELECT COUNT(*) FROM responses WHERE option_id = 1 AND poll_id = 1 ) /
          (SELECT COUNT(*) FROM responses WHERE poll_id = 1) AS all_percent
          FROM responses
          GROUP BY user_id
         ) t
    ORDER BY deviation DESC

Answer 2

您需要使用变量@var：=

例如......

... SELECT @var：= COUNT（*）FROM answers WHERE poll_id = 1）AS all_percent， ABS（@ var1 - @ var2）AS偏差......

Answer 3

SELECT * 
FROM   (SELECT user_id, 
               (SELECT COUNT(1) 
                FROM   responses 
                WHERE  option_id = 1 
                       AND poll_id = 1 
                       AND pref = 1) / (SELECT COUNT(1) 
                                        FROM   responses 
                                        WHERE  poll_id = 1 
                                               AND pref = 1)   AS pref_percent, 
               (SELECT COUNT(1) 
                FROM   responses 
                WHERE  option_id = 1 
                       AND poll_id = 1) / (SELECT COUNT(1) 
                                           FROM   responses 
                                           WHERE  poll_id = 1) AS all_percent, 
               Abs(all_percent - pref_percent)                 AS deviation 
        FROM   responses 
        GROUP  BY user_id) t 
ORDER  BY t.deviation DESC

MySQL：按用户定义列的绝对值排序

3 个答案: