使用mysql根据用户的最高得分计算其排名 [英] Calculate rank of the users based on their max score using mysql

问题描述

我有一张表格(称为用户)，我需要根据他们的得分对用户进行排名，但我希望根据用户的最高得分进行排名.

I have a table (called users) I need rank of users based on their score but I want rank on the bases of users max score.

+-----------+------------+
| User_id   | Score      |
+-----------+------------+
| 1         | 12258      | 
| 1         | 112        |
| 2         | 9678       |
| 5         | 9678       |
| 3         | 689206     |
| 3         | 1868       |

期望结果

+-----------+------------+---------+
| User_id   | Score      | Rank    |
+-----------+------------+---------+
| 3         | 689206     |   1     |     
| 1         | 12258      |   2     |
| 2         | 9678       |   3     |
| 5         | 9678       |   3     |

推荐答案

您正在寻找DENSE_RANK，但它支持高于8.0的mysql版本

You are looking for DENSE_RANK, But it supports mysql version higher than 8.0

1. 使用相关子查询按每个User_id来获取max值
2. 使用两个变量存储一个rank，另一个存储先前的值以生成DENSE_RANK.

1. use correlated-subquery to get max value by each User_id
2. use two variables one to store rank another to store previous value to make the DENSE_RANK number.

看起来像这样.

CREATE TABLE T(
   User_id int,
   Score int
); 

insert into t values (1,12258); 
insert into t values (1,112);
insert into t values (2,9678);
insert into t values (5,9678);
insert into t values (3,689206);
insert into t values (3,1868);

查询1 :

SELECT User_id,Score,Rank
FROM (
  SELECT User_id,
         Score,
         @rank :=IF(@previous = t1.score, @rank, @rank + 1) Rank,
         @previous := t1.Score
  FROM T t1 CROSS JOIN (SELECT @Rank := 0,@previous := 0) r
  WHERE t1.Score = 
  (
    SELECT  MAX(Score)  
    FROM T tt
    WHERE t1.User_id = tt.User_id
  ) 
  ORDER BY Score desc
) t1

结果 :

Results:

| User_id |  Score | Rank |
|---------|--------|------|
|       3 | 689206 |    1 |
|       1 |  12258 |    2 |
|       2 |   9678 |    3 |
|       5 |   9678 |    3 |

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