MySQL5.7中计算密集秩的另一个技巧（如MySQL8中）是在使用变量赋值时使用case。

SELECT User_id, MaxScore AS Score,
  CASE 
  WHEN MaxScore = @prevScore THEN @rnk
  WHEN @prevScore := MaxScore THEN @rnk := @rnk+1 
  ELSE @rnk := @rnk+1
  END AS Rank
FROM 
(
  SELECT User_id, MAX(Score) AS MaxScore
  FROM YourTable
  GROUP BY User_id
  ORDER BY MaxScore DESC, User_id
) AS q
CROSS JOIN (SELECT @rnk := 0, @prevScore := null) AS vars

你可以在rextester上测试。

你在找什么 DENSE_RANK ，但支持8.0以上的mysql版本
使用相关子查询获取 max 按每个值 User_id 使用两个变量一个存储 rank 另一个用于存储以前的值以使 DENSE_RANK 号码。
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：

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