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`));
输入描述:
无
输出描述:
emp_no |
salary |
rank |
---|---|---|
10005 |
94692 |
1 |
10009 |
94409 |
2 |
10010 |
94409 |
2 |
10001 |
88958 |
3 |
10007 |
88070 |
4 |
10004 |
74057 |
5 |
10002 |
72527 |
6 |
10003 |
43311 |
7 |
10006 |
43311 |
7 |
10011 |
25828 |
8 |
思路:题目给的输出跟题目描述不符,输出是按照salary降序排列的。所以写代码时还是得根据输出来写。
SELECT
s1.emp_no,
s1.salary,
count(DISTINCT s2.salary) AS rank
FROM
salaries s1,
salaries 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;