https://www.nowcoder.com/practice/c1472daba75d4635b7f8540b837cc719?tpId=82&tags=&title=&difficulty=0&judgeStatus=0&rp=1
题解来自:https://blog.nowcoder.net/n/f35b41269fd84707a748724827510e23?f
方法一好理解
select s.emp_no, s.salary, e.last_name, e.first_name from salaries s join employees e on s.emp_no = e.emp_no where s.salary = -- 第三步: 将第二高工资作为查询条件 ( select max(salary) -- 第二步: 查出除了原表最高工资以外的最高工资(第二高工资) from salaries where salary < ( select max(salary) -- 第一步: 查出原表最高工资 from salaries where to_date = '9999-01-01' ) and to_date = '9999-01-01' ) and s.to_date = '9999-01-01'
方法二:
select s.emp_no, s.salary, e.last_name, e.first_name from salaries s join employees e on s.emp_no = e.emp_no where s.salary = ( select s1.salary from salaries s1 join salaries s2 -- 自连接查询 on s1.salary <= s2.salary group by s1.salary -- 当s1<=s2链接并以s1.salary分组时一个s1会对应多个s2 having count(distinct s2.salary) = 2 -- (去重之后的数量就是对应的名次) and s1.to_date = '9999-01-01' and s2.to_date = '9999-01-01' ) and s.to_date = '9999-01-01'