Question

我有一个结果表，如：


ID  STUDENT_ID  Branch_id   Class_id    Exam_id Subject_id  Numbers     Date
1       653         5           1           1   8               60      2012-01-01
2       653         5           1           1   9               40      2012-01-01
3       653         5           1           1   10              80      2012-01-01
4       653         5           1           1   11              50      2012-01-01
5       653         5           1           1   12              65      2012-01-01
6       653         5           1           1   13              33      2012-01-01
7       653         5           1           1   15              86      2012-01-01
8       222         5           1           1   8               100     2012-01-01
9       222         5           1           1   9               80      2012-01-01
10      222         5           1           1   10              92      2012-01-01
11      222         5           1           1   11              50      2012-01-01
12      222         5           1           1   12              65      2012-01-01
13      222         5           1           1   13              33      2012-01-01
7       222         5           1           1   15              86      2012-01-01

我的欲望结果如：


Student_ID  Math    English      Science    Total   Rank
1       90  89      88  267  1
2       90  89      88  267  1
3       58  45      98  201      2

我希望通过这种方法获得学生排名：

Reference Link

SET @rank = 0, @prev_val = NULL;

SELECT rank, correct FROM 
(
    SELECT 
        @rank := IF(@prev_val=correct,@rank,@rank+1) AS rank,
        @prev_val := correct AS correct, uid
    FROM quiz_user 
    ORDER BY correct DESC
)as result WHERE uid=xxxxxxxxxxxx

这个查询我只需要表结构之间的差异，作者的帖子在正确的列上分配一个等级，我需要在数字SUM（数字）列之后对所有数字进行排序。

请帮助。

Answer 1

试试这个：

SELECT STUDENT_ID, Numbers, IF(@marks=(@marks:=Numbers), @auto, @auto:=@auto+1) rank 
FROM (SELECT STUDENT_ID, SUM(Numbers) Numbers
      FROM quiz_user 
      GROUP BY STUDENT_ID 
      ORDER BY Numbers DESC, STUDENT_ID 
     ) AS A, (SELECT @auto:=0, @marks:=0) AS B;

通过Mysql查询动态获取排名

1 个答案: