Fetch the repository succeeded.
按照salary的累计和running_total，其中running_total为前两个员工的salary累计和，其他以此类推。 具体结果如下Demo展示。
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 s2.emp_no,s2.salary,sum(s1.salary) running_total from (select emp_no,salary from salaries where to_date = '9999-01-01') s1 inner join (select emp_no,salary from salaries where to_date = '9999-01-01') s2 on s1.emp_no <= s2.emp_no group by s2.emp_no1、获得当前员工的工资情况s1
select emp_no,salary from salaries where to_date = '9999-01-01'2、连接当前员工信息s2，筛选出emp_no小于当前用户的所有员工信息
s1 inner join (select emp_no,salary from salaries where to_date = '9999-01-01') s2 on s1.emp_no <= s2.emp_no3、根据emp_no进行分组，对salary列进行求和
select s2.emp_no,s2.salary,sum(s1.salary) running_total from s1 inner join s2 on s1.emp_no <= s2.emp_no group by s2.emp_noSign in for post a comment
Comment ( 0 )