题目描述：

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    |+-------+------+

SQL架构：

Create table If Not Exists Scores (Id int, Score DECIMAL(3,2));Truncate table Scores;insert into Scores (Id, Score) values ('1', '3.5');insert into Scores (Id, Score) values ('2', '3.65');insert into Scores (Id, Score) values ('3', '4.0');insert into Scores (Id, Score) values ('4', '3.85');insert into Scores (Id, Score) values ('5', '4.0');insert into Scores (Id, Score) values ('6', '3.65');

解题思路：

　　oracle可以使用排名函数进行排名，因为名次之间不应该有“间隔”，所以要使用dense_rank()函数，这个函数排的名次是连续的，rank()函数会跳过（并列排名），mysql依然是使用自定义变量来进行排名。

解题方案：

　　oracle

select round(a.score,2) as score, b.rank  from Scores a  join (select a.*, rownum as rank          from (select distinct a.score from Scores a order by a.score desc) a) b    on a.score = b.score order by b.rank

　　mysql

SELECT    ROUND(a.score, 2) AS score,    b.rankFROM    Scores aJOIN (    SELECT        a.*, (@rowNum :=@rowNum + 1) AS rank    FROM        (            SELECT DISTINCT                a.score            FROM                Scores a        ) a,        (SELECT(@rowNum := 0)) b    ORDER BY        a.score DESC) b ON a.score = b.scoreORDER BY    b.rank