例：查找当前薪水(to_date=‘9999-01-01‘)排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
解答思路：
此题方法较多
方法一：
利用两次MAX，连同select嵌套
主要思想为多层SELECT嵌套与MAX()函数结合
1、先利用MAX()函数找出salaries中当前薪水最高者，即SELECT MAX(salary) FROM salaries WHERE to_date = ‘9999-01-01‘
2、再利用INNER JOIN连接employees与salaries表，限定条件为【同一员工】e.emp_no = s.emp_no、【当前】s.to_date = ‘9999-01-01‘与【非薪水最高】s.salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = ‘9999-01-01‘)
3、在以上限制条件下找薪水最高者，即为所有员工薪水的次高者
SELECT e.emp_no, MAX(s.salary) AS salary, e.last_name, e.first_name FROM employees AS e INNER JOIN salaries AS s ON e.emp_no = s.emp_noWHERE s.to_date = ‘9999-01-01‘AND s.salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = ‘9999-01-01‘)方法二:
利用max选出最大，用where 限定小于最大值的，然后再取最大的那个
select a.emp_no,max(b.salary),a.last_name,a.first_name
from employees a inner join salaries b
on a.emp_no=b.emp_no
where b.salary<(select max(salary) from salaries)
and b.to_date = ‘9999-01-01‘
