MySQL 分组取前几示例：部门工资前三高的员工

这是个有点难度而且在工作中有用到的 SQL。

要求：编写一个SQL，获取部门工资前三高的员工。

员工表和部门表结构：

CREATE TABLE `employee` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255),
  `salary` decimal(10,2),
  `department_id` int(11),
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

CREATE TABLE `department` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255),
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

员工表和部门表数据：

INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (1, 'Joe', 70000.00, 1);
INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (2, 'Henry', 80000.00, 2);
INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (3, 'Sam', 60000.00, 2);
INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (4, 'Max', 90000.00, 1);
INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (5, 'Janet', 69000.00, 1);
INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (6, 'Randy', 85000.00, 1);
INSERT INTO `employee`(`id`, `name`, `salary`, `department_id`) VALUES (7, 'Eva', 85000.00, 1);

INSERT INTO `department`(`id`, `name`) VALUES (1, 'IT');
INSERT INTO `department`(`id`, `name`) VALUES (2, 'Sales');

题库的答案：

SELECT
	d.`name` AS '部门',
	e.`name` AS '员工',
	e.salary AS '工资' 
FROM
	employee e
	JOIN department d ON d.id = e.department_id 
WHERE
	(
	SELECT count(DISTINCT em.salary) FROM employee em WHERE em.salary > e.salary AND em.department_id = e.department_id
	) < 3 
ORDER BY e.department_id, e.salary DESC

输出结果如下：

部门	员工	工资
IT	Max	90000
IT	Randy	85000
IT	Eva	85000
IT	Joe	70000
Sales	Henry	80000
Sales	Sam	60000

首先来理解一下上面的 SQL，当 < 3 的条件改为 = 0 时，即子表中相同部门没有比主表工资高的员工，则取得工资最高的员工；当条件为 = 1 时，表示子表中相同部门里只有一个比主表工资高的员工，则取得工资第二高的员工；同理，条件 = 2 表示工资第三高的员工，所以工资前三高的员工的条件为 < 3。

通过结果可以看到，第二名员工和第三名员工工资相同，被当作并列第二，并不会排挤掉第三名。如果我们希望出现并列第二名时，第三名就变成第四名呢？可以把 count(DISTINCT em.salary) 改成 count(*)。

SELECT
	d.`name` AS '部门',
	e.`name` AS '员工',
	e.salary AS '工资' 
FROM
	employee e
	JOIN department d ON d.id = e.department_id 
WHERE
	(
	SELECT count(*) FROM employee em WHERE em.salary > e.salary AND em.department_id = e.department_id
	) < 3 
ORDER BY e.department_id, e.salary DESC

输出结果：

部门	员工	工资
IT	Max	90000
IT	Randy	85000
IT	Eva	85000
Sales	Henry	80000
Sales	Sam	60000

上面的写法中，当我们取前两名时，会得到 IT 部门的第一名和两个第二名的员工。如果我们希望去掉并列的情况，即就算工资相同也分为不同名次呢？那可以根据工资排序来增加多一个序号列，把 employee 表替换成下面这个子表：

SELECT (@i:=@i+1) AS rownum, es.* FROM employee es, (select @i:=0) ri ORDER BY es.salary

然后去查询每个部门工资前两名的员工，这里注意一下，两个子表变量名需要不一样：

SELECT
	d.`name` AS '部门',
	e.`name` AS '员工',
	e.salary AS '工资' 
FROM
	(SELECT (@i:=@i+1) AS rownum, es.* FROM employee es, (select @i:=0) ri ORDER BY es.salary) e
	JOIN department d ON d.id = e.department_id 
WHERE
	(
	SELECT count(*) FROM (SELECT (@j:=@j+1) AS rownum, es.* FROM employee es, (select @j:=0) rj ORDER BY es.salary) em WHERE em.rownum > e.rownum AND em.department_id = e.department_id
	) < 2
ORDER BY e.department_id, e.salary DESC

结果如下：

部门	员工	工资
IT	Max	90000
IT	Randy	85000
Sales	Henry	80000
Sales	Sam	60000

参考：部门工资前三高的员工 - 力扣 (LeetCode)