Mysql 从SQL中检索最新日期并获取差异

我的桌子是：

QuestionMaster

预期产出为：

+-----+-----------------+
|qId |差异日|
+-----+-----------------+
|1    |6                |
+-----+-----------------+
|2    |23               |
+-----+-----------------+
|3    |28               |
+-----+-----------------+

有什么建议吗？
提前感谢。
您需要获取
MAX（answerDate）
，否则MySQL将使用每个组中某些不确定记录的值

SELECT   q.qId, DATEDIFF(MAX(a.answerDate), q.quesDate)
FROM     QuestionMaster q JOIN AnswerMaster a ON a.quesId = q.qId
GROUP BY q.qId

请参见。
如果我在这种情况下使用GROUP BY，我不知道会发生什么“我不想使用子查询？”你是认真的吗？您选择了一种以某种方式工作的语言和技术，但您不想使用它？你有什么理由吗？这只是我实际询问的一部分。这就是为什么可能存在性能问题。
+--------+------+---------------------+
|answerId|quesId|answerDate           |
+--------+------+---------------------+
|1       |1     |2014-02-08 11:10:05  |
+-----------+------+------------------+
|2       |1     |2014-02-10 18:06:05  |
+--------+------+---------------------+
|3       |1     |2014-02-07 12:07:10  |
+--------+------+---------------------+
|4       |2     |2014-02-09 09:18:05  |
+--------+------+---------------------+
|5       |2     |2014-03-01 08:06:05  |
+--------+------+---------------------+
|6       |2     |2014-02-15 16:10:45  |
+--------+------+---------------------+
|7       |2     |2014-02-12 06:35:25  |
+--------+------+---------------------+
|8       |3     |2014-03-05 08:09:05  |
+--------+------+---------------------+
|9       |3     |2014-02-25 15:02:19  |
+--------+------+---------------------+
SELECT 
question.qid as questionId,
timestampdiff(DAY,question.quesDate,answer.answerDate) AS `answerTime`
FROM QuestionMaster question
Left Join AnswerMaster answer on question.qid = answer.quesid
group by question.qid ;

+-----+-----------------+
|qId  |DifferenceInDays |
+-----+-----------------+
|1    |6                |
+-----+-----------------+
|2    |23               |
+-----+-----------------+
|3    |28               |
+-----+-----------------+
SELECT   q.qId, DATEDIFF(MAX(a.answerDate), q.quesDate)
FROM     QuestionMaster q JOIN AnswerMaster a ON a.quesId = q.qId
GROUP BY q.qId