题目描述
对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名,相同salary并列且按照emp_no升序排列
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`));
解法:
1. MySQL
select s1.emp_no , s1.salary,count(distinct s2.salary) as rank
from salaries as s1,salaries as s2
where s1.to_date = '9999-01-01'
and s2.to_date = '9999-01-01'
and s1.salary <= s2.salary
group by s1.emp_no
order by s1.salary desc,s1.emp_no asc
思路:
1、count的是distinct
2、重点是 s1.salary <= s2.salary group by s1.emp_no
这样count就是rank排名,是在输出s1.salary的情况下,有多少个s2.salary大于等于s1.salary
2. hadoop(SQL)
select emp_no , salary,DESEN_RANK() OVER(ORDER BY salary desc,emp_no asc) as rank
from salaries where to_date = '9999-01-01'