mysql - 选择中具有多个子查询的上表可见性

我有一个查询：

SELECT b.user_id, b.active FROM users b
WHERE b.followers_count != (SELECT COUNT(*)
    FROM (SELECT u.user_id
        FROM user_follow uf,users u,user_follow_request ufr
        WHERE
        uf.following_id = b.user_id AND
        uf.following_id = ufr.friend_id AND
        ufr.status = 'approved' AND
        ufr.user_id = u.user_id AND u.user_id != b.user_id AND u.active != 0
        GROUP BY u.user_id) a)
 AND b.active = -1 limit 5;

它必须从user_id中选择与sql计算的列中具有不同users的所有followers_count。

但是问题是我收到错误消息

Error Code: 1054. Unknown column 'b.user_id' in 'on clause'

我做错了什么？非常感谢您的帮助。

最佳答案

您是说uf.user_id = b.user_id而不是uf.following_id = b.user_id吗？无论如何，user_follow表用于什么？

同意Strawberry，您最好使用显式联接。它们更容易阅读和理解。这是您可能已经提到的内容（请注意，根本不使用user_follow表）：

SELECT u.user_id, u.followers_count, IF(ISNULL(uafc.actual_fc),0,uafc.actual_fc) AS afc
  FROM users AS u
    LEFT JOIN (
      SELECT u.user_id AS user_id, COUNT(*) AS actual_fc
        FROM users AS u
          JOIN user_follow_request AS ufr
            ON ufr.friend_id = u.user_id
          JOIN users AS fu
            ON fu.user_id = ufr.user_id
        WHERE ufr.user_id != u.user_id
          AND ufr.status = 'approved'
          AND fu.active != 0
        GROUP BY u.user_id, u.followers_count
      ) AS uafc
      ON uafc.user_id = u.user_id
  WHERE u.followers_count != IF(ISNULL(uafc.actual_fc),0,uafc.actual_fc)
;

可能有一些更好的NULL解决方法。

在SQLFiddle上查看：http://sqlfiddle.com/#!2/71f6c/3