MySQL 排名函数 RANK, DENSE_RANK, ROW_NUMBER

1 排名函数有哪些?

• RANK(): 并列跳跃排名, 并列即相同的值, 相同的值保留重复名次, 遇到下一个不同值时, 跳跃到总共的排名
• DENSE_RANK(): 并列连续排序, 并列即相同的值, 相同的值保留重复名次, 遇到下一个不同值时, 依然按照连续数字排名
• ROW_NUMBER(): 连续排名, 即使相同的值, 依旧按照连续数字进行排名

RANK, DENSE_RANK, ROW_NUMBER 函数在查询结果上的区别

2 SQL 代码实现

t_scores 表

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| score       | decimal |
+-------------+---------+

在 SQL 中, id 是该表的主键
该表的每一行都包含了一场比赛的分数, score 是一个有两位小数点的浮点值

2.1 RANK

RANK(): 并列跳跃排名, 并列即相同的值, 相同的值保留重复名次, 遇到下一个不同值时, 跳跃到总共的排名
RANK() OVER(ORDER BY score DESC) 对 score 字段升序排序, 进行排名

SELECT score,  RANK() OVER(ORDER BY score ASC) as "rank" FROM t_scores;

查询结果:

2.2 DENSE_RANK

DENSE_RANK(): 并列连续排序, 并列即相同的值, 相同的值保留重复名次, 遇到下一个不同值时, 依然按照连续数字排名

SELECT score,  DENSE_RANK() OVER(ORDER BY score ASC) as "rank" FROM t_scores;

查询结果:

2.3 ROW_NUMBER

ROW_NUMBER(): 连续排名, 即使相同的值, 依旧按照连续数字进行排名

SELECT score,  ROW_NUMBER() OVER(ORDER BY score ASC) as "rank"  FROM t_scores;

查询结果:

2.4 OVER

函数语法

OVER (PARTITION BY (expr) ORDER BY (expr))

LeetCode–185 部门工资前三高的所有员工

思路: 按照部门统计, 前三高薪水的员工信息, 需要用到排名函数, 需要保持排名的连续性, 结合 partition by e.departmentId 按部门统计

dense_rank() over(partition by e.departmentId order by e.salary desc)

Employee 表:

+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+

Department 表:

+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+

输出:

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

最终SQL

select ed.Department, ed.Employee, ed.Salary
from (select d.name as Department, e.name as Employee, e.salary as Salary,
			dense_rank() over(partition by e.departmentId order by e.salary desc) as denseRank 
	from Employee as e inner join Department as d on e.departmentId = d.id) ed
where ed.denseRank <= 3