Fetch the repository succeeded.
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`));select t1.emp_no,t1.salary,count(distinct t2.salary) rank from (select * from salaries s where to_date = '9999-01-01') t1 , (select * from salaries s where to_date = '9999-01-01') t2 where t2.salary >= t1.salary group by t1.emp_no order by rank本题无法过MySQL的查询，但是测试通过了sqlite，也就是牛客网提交通过。
1、获得两个临时表t1,t2
(select * from salaries s where to_date = '9999-01-01') t1
(select * from salaries s where to_date = '9999-01-01') t22、用第一个临时表的t1.salary去和t2.salary进行比较，从何可以获得所有t1.salary小于t2.salary的数据集，正是如此，我们可以对用户进行分组，统计每个用户对应的salary小于其他用户的数据条数，此为排名。为什么count时要去重，因为可能存在相同排名的人
select t1.emp_no,count(distinct t2.salary) from t1,t2 where t2.salary >= t1.salary group by t1.emp_noSign in for post a comment
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