我有与此问题中描述的非常类似的要求。
Rank users in mysql by their points
唯一的区别在于我的数据。上述问题的数据表中每个学生只有一行。但在我的情况下,可能有一个表包含多个行,对于像这样的单个学生
所以现在排名应该根据用户拥有的SUM
点数(总计)来计算。所以在这种情况下结果将是。
我尝试了上述问题的解决方案,但无法得到结果。任何帮助将不胜感激。
答案 0 :(得分:4)
在我之前的回答中使用相同的查询只需更改表学生的子查询,以组合每个学生的所有记录
change [student er] for
(SELECT `id`, SUM(`points`) as `points`
FROM students
GROUP BY `id`) er
select er.*,
(@rank := if(@points = points,
@rank,
if(@points := points,
@rank + 1,
@rank + 1
)
)
) as ranking
from (SELECT `id`, SUM(`points`) as `points`
FROM students
GROUP BY `id`) er cross join
(select @rank := 0, @points := -1) params
order by points desc;
输出
| id | points | ranking |
|----|--------|---------|
| 5 | 91 | 1 |
| 6 | 81 | 2 |
| 1 | 80 | 3 |
| 2 | 78 | 4 |
| 3 | 78 | 4 |
| 4 | 77 | 5 |
| 7 | 66 | 6 |
| 8 | 15 | 7 |
答案 1 :(得分:3)
试试这个:
select id, points, @row := ifnull(@row, 0) + diff rank
from (select *, ifnull(@prev, 0) != points diff, @prev := points
from (select id, sum(points) points
from students
group by 1
order by 2 desc) x) y
请参阅SQLFiddle
答案 2 :(得分:1)
EDITED: (这应该有用)
SELECT I.Id, I.Points, Rk.Rank
FROM
(SELECT Id, Points, @Rk := @Rk+1 As Rank
FROM (SELECT id, SUM(points) AS Points
FROM students
GROUP BY id
ORDER BY Points DESC) As T,
(SELECT @Rk := 0) AS Rk) As I
INNER JOIN
(SELECT *
FROM (
SELECT Id, Points, @Rk2 := @Rk2+1 As Rank
FROM (SELECT id, SUM(points) AS Points
FROM students
GROUP BY id
ORDER BY Points DESC) As T1,
(SELECT @Rk2 := 0) AS Rk) AS T2
GROUP BY Points) As Rk
USING(Points)
输出将是:
| Id | Points | Rank |
|----|--------|---------|
| 5 | 91 | 1 |
| 6 | 81 | 2 |
| 1 | 80 | 3 |
| 2 | 78 | 4 |
| 3 | 78 | 4 |
| 4 | 77 | 6 |
| 7 | 66 | 7 |
| 8 | 15 | 8 |
在第四个位置的两个Ids之后,你将获得第六个位置,因为5个Ids在第六个位置之前。