这题需要注意
无
| 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 |
sql:
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
SQL-23 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
原文:https://www.cnblogs.com/kexiblog/p/10684265.html