Question

我希望GROUP BY选择最早的StartSessionTime然后我想基于StartSessionTime进行过滤，我尝试了各种各样的东西，最近的是：

$sql_d = "SELECT  a.* FROM Session_Log  a INNER JOIN 
     ( SELECT  ID, MIN(SessionStartTime) 
        FROM Session_Log GROUP BY ID 
     ) b 
     ON a.ID = b.ID 
        AND a.SessionStartTime = b.SessionStartTime 
     WHERE DATE(SessionStartTime) < ( '2018-01-01' + INTERVAL $b DAY ) 
      AND  DATE(SessionStartTime) >= ( '2018-01-01' + INTERVAL $a DAY )";

请帮助正确表达

Answer 1

您需要为MIN(SessionStartTime)指定别名，以便在ON子句中引用它。

SELECT a.*
FROM FMan_Session_Log AS a
JOIN (
    SELECT DeviceID, MIN(SessionStartTime) AS MinSessionStartTime
    FROM FMan_Session_Log
) AS b ON a.DeviceID = b.DeviceID AND a.SessionStartTime = b.MinSessionStartTime
WHERE a.SessionStartTime >= DATE_ADD('2018-01-01', INTERVAL $a DAY)
AND a.SessionStartTime < DATE_ADD('2018-01-01, INTERVAL $b DAY)

Answer 2

修正了错误。首先没有为b定义SessionStartTime。其次，SessionStartTime在WHERE子句

中是不明确的

$sql_d = "SELECT  a.* FROM Session_Log  a INNER JOIN 
     ( SELECT  ID, MIN(SessionStartTime) **as SessionStartTime**
        FROM Session_Log GROUP BY ID 
     ) b 
     ON a.ID = b.ID 
        AND a.SessionStartTime = b.SessionStartTime 
     WHERE DATE(a.SessionStartTime) < ( '2018-01-01' + INTERVAL $b DAY ) 
      AND  DATE(a.SessionStartTime) >= ( '2018-01-01' + INTERVAL $a DAY )";

Mysql GROUP BY首先记录然后过滤该记录

2 个答案: