SQL count与distinct的结合使用

版权声明:本文为博主原创文章,未经博主允许不得转发,博客地址: https://blog.csdn.net/weixin_41180516 https://blog.csdn.net/weixin_41180516/article/details/83903751

编写一个 SQL 查询来实现分数排名。如果两个分数相同,则两个分数排名(Rank)相同。请注意,平分后的下一个名次应该是下一个连续的整数值。换句话说,名次之间不应该有“间隔”。

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

例如,根据上述给定的 Scores 表,你的查询应该返回(按分数从高到低排列):

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

select Score,(select count(distinct score) from Scores where score >= s.score) as Rank from Scores s order by score desc

distinct score 去重

{"headers":["Score","Rank"],"values":[[4.0,1],[4.0,1],[3.85,2],[3.65,3],[3.65,3],[3.5,4]]}

猜你喜欢

转载自blog.csdn.net/weixin_41180516/article/details/83903751