Question

我正在尝试计算freelanceFeedback的顺序和计数顺序如下：

$sql = "SELECT authentication.*, (SELECT COUNT(*) FROM freelanceFeedback) as taskscount FROM authentication
            LEFT JOIN freelanceFeedback
            ON authentication.userId=freelanceFeedback.FK_freelanceWinnerUserId
            WHERE `FK_freelanceProvider`=$what
            ORDER BY taskscount DESC";

但是如果用户有多个反馈并且没有通过taskscount进行排序，那么我有多个输出。

我无法弄清楚'推文'是错误的......

**更新** 我想我自己也有：

$sql = "SELECT DISTINCT authentication.*, 
            (SELECT COUNT(*) FROM freelanceFeedback
            WHERE FK_freelanceWinnerUserId=userId
            ) as taskscount 
            FROM authentication
            WHERE `FK_freelanceProvider`=$what
            ORDER BY taskscount DESC";

这只输出1个用户并按反馈量进行订购。

Answer 1

使用COUNT（）时，还需要使用GROUP BY：

    SELECT authentication.userId, 
           COUNT(freelanceFeedback.id) AS taskscount 
      FROM authentication
 LEFT JOIN freelanceFeedback
        ON authentication.userId = freelanceFeedback.FK_freelanceWinnerUserId
     WHERE `FK_freelanceProvider`= $what
  GROUP BY authentication.userId
  ORDER BY taskscount DESC

但是，这只有在你没有做SELECT *时才会起作用（无论如何这都是不好的做法）。不在COUNT位中的所有内容都需要进入GROUP BY。如果这包括文本字段，您将无法执行此操作，因此您需要对子查询执行JOIN操作。如果不这样做，MySQL不会抱怨，但它会严重减慢速度，而其他数据库会抛出错误，所以最好这样做：

    SELECT authentication.userId, 
           authentication.textfield, 
           authentication.othertextfield,
           subquery.taskscount
      FROM authentication
 LEFT JOIN (SELECT freelanceFeedback.FK_freelanceWinnerUserId,
                   COUNT(freelanceFeedback.FK_freelanceWinnerUserId) AS taskscount 
              FROM freelanceFeedback
          GROUP BY FK_freelanceWinnerUserId) AS subquery
        ON authentication.userId = subquery.FK_freelanceWinnerUserId
     WHERE authentication.FK_freelanceProvider = $what
  ORDER BY subquery.taskscount DESC

目前还不清楚FK_freelanceProvider是哪一个表，所以我假设它是身份验证。

MySQL - 依靠连接

1 个答案: