题目描述
获取当前（to_date='9999-01-01'）薪水第二多的员工的emp_no以及其对应的薪水salary 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`));
输入描述:
无
输出描述:
emp_nosalary1000994409
SELECT S.emp_no,S.salary
FROM salaries S
WHERE S.salary=(
SELECT DISTINCT S.salary
FROM salaries S
WHERE S.to_date = '9999-01-01'
GROUP BY S.salary DESC
LIMIT 1 OFFSET 1
);
--法2：row_number开窗函数(当薪水第二的人有多个时只会输出一个)
select emp_no,salary
from
(select emp_no,salary,row_number() over(order by salary desc) t
from salaries
where to_date='9999-01-01')
where t=2
--法3：rank开窗函数(当薪水第二的人有多个时会全部输出)
select emp_no,salary
from
(select emp_no,salary,rank() over(order by salary desc) t
from salaries
where to_date='9999-01-01')
where t=2
--法4：max()+子查询
select emp_no,max(salary)
from salaries
where salary <>(select max(salary)
from salaries
where to_date='9999-01-01')
