SQL以获取group by的最大日期

Question

我有一个像这样的交易表......

-------------------------------------------------------------------------------------
Subject_id | Question_id | Answerd_id | Student_id | Answer_text | Insert_DTM
-------------------------------------------------------------------------------------
5005       | 3004        | 1004       | 1309       | test        | 2018-05-31 12:07:42
-------------------------------------------------------------------------------------
5005       | 3005        | 1005       | 1309       | test        | 2018-05-31 12:07:42
-------------------------------------------------------------------------------------
5005       | 3004        | NULL       | 1309       | Null        | 2018-05-31 12:09:43
-------------------------------------------------------------------------------------
5005       | 3002        | NULL       | 1309       | Null        | 2018-05-31 12:07:42
-------------------------------------------------------------------------------------
5005       | 3005        | 1005       | 1309       | test        | 2018-05-31 11:35:47
-------------------------------------------------------------------------------------
5005       | 3005        | 1005       | 1309       |             |2018-05-31 11:36:37

从这张表中我必须找出一个学生和主题id最近回答的行。

我正在使用的SQL ..

SELECT subject_id, 
       question_id, 
       answer_id, 
       student_id, 
       answer_text, 
       insert_dtm 
FROM   exam_trans 
WHERE  student_id = 1309 
       AND subject_id = 5005 
       AND insert_dtm IN (SELECT Max(insert_dtm) 
                          FROM   exam_trans 
                          WHERE  student_id = 1309 
                                 AND subject_id = 5005 
                          GROUP  BY question_id) 
GROUP  BY subject_id, 
          question_id, 
          answer_id, 
          member_id, 
          answer_text, 
          insert_dtm 
ORDER  BY answer_id, 
          question_id DESC 

但作为输出我得到了

-------------------------------------------------------------------------------------
Subject_id | Question_id | Answerd_id | Student_id | Answer_text | Insert_DTM
-------------------------------------------------------------------------------------
5005       | 3004        | 1004       | 1309       | test        | 2018-05-31 12:07:42
-------------------------------------------------------------------------------------
5005       | 3005        | 1005       | 1309       | test        | 2018-05-31 12:07:42
---------------------------------------------------------------------------------------
5005       | 3004        | NULL       | 1309       | Null        | 2018-05-31 12:09:43
--------------------------------------------------------------------------------------
5005       | 3002        | NULL       | 1309       | Null        | 2018-05-31 12:07:42

---

3004在输出中出现两次，但我的预期输出是

---

Subject_id | Question_id | Answerd_id | Student_id | Answer_text | Insert_DTM
-------------------------------------------------------------------------------------
5005       | 3005        | 1005       | 1309       | test        | 2018-05-31 12:07:42
---------------------------------------------------------------------------------------
5005       | 3004        | NULL       | 1309       | Null        | 2018-05-31 12:09:43
--------------------------------------------------------------------------------------
5005       | 3002        | NULL       | 1309       | Null        | 2018-05-31 12:07:42

---

3004应该只有一个最近的时间戳只....

任何人都可以帮助我正确的SQL ...

我在AWS中使用Oracle RDS

Answer 1

为此更好用ROW_NUMBER()

SELECT *
FROM (SELECT *,
             ROW_NUMBER() OVER (PARTION BY Question_id, Subject_id
                                ORDER BY Insert_DTM DESC) as rn
      FROM exam_trans 
      WHERE  student_id = 1309 
        AND  subject_id = 5005 
     )
WHERE rn = 1

Answer 2

你似乎想要：

select e.*
from exam_trans e
where Insert_DTM = (select max(e1.Insert_DTM)
                    from  exam_trans e1
                    where e1.Subject_id = e.Subject_id and
                          e1.Question_id = e.Question_id
                   );

Answer 3

您可以使用子查询和row_number（）。

https://stackoverflow.com/a/10515397/7328059应该回答你的问题。