标签：from style 定义变量 exists 连续 esc sel scores 应该
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 | +-------+------+
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‘);
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。
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.rank FROM Scores a JOIN ( 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.score ORDER BY b.rank
标签：from style 定义变量 exists 连续 esc sel scores 应该
原文地址：https://www.cnblogs.com/zouqf/p/10283410.html
