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
1
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
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