Mysql SUM查询未返回预期结果

时间:2015-10-13 01:01:48

标签: php mysql sql

您好我寻求帮助以查看我的查询有什么问题,该查询没有返回表consulta中行的计数(数字),其中userIDC等于来自表用户的userID。以及来自表asesoria的userIDC的相同操作。

SELECT c.*, SUM(IF(a.userIDA = c.userID , 1, 0)) AS count_asesoria, SUM(IF(s.userIDC = c.userID , 1, 0)) AS count_consulta
   FROM users as c 
   LEFT JOIN consulta AS s ON s.userIDC = c.userID 
   LEFT JOIN asesoria AS a ON a.userIDA = c.userID
   GROUP BY c.userID DESC 

现在它为count_asesoriacount_consulta

重复了相同的结果

表用户:

userID | Data    |
------------------
3      | content |
表咨询

userIDC | Data   |
------------------
3      | content |
3      | content |

所以count_consulta必须返回2

1 个答案:

答案 0 :(得分:0)

问题的最可能原因是连接中表之间的笛卡尔积。正确的解决方案是在查询之前预先聚合结果。或者,使用相关子查询。在这种情况下,这可能是最简单的方法,并且可能具有最佳性能:

SELECT u.*,
       (SELECT COUNT(*)
        FROM consulta c
        WHERE c.userIDC = u.userId
       ) as count_consulta, 
       (SELECT COUNT(*)
        FROM asesoria a
        WHERE a.userIDA = u.userId
       ) as count_asesoria
FROM users u;

为获得最佳效果,您需要consulta(userIDA)asesoria(userIDA)上的索引。