前N个员工的salary累计和

------------恢复内容开始------------

一、问题

按照salary的累计和running_total，其中running_total为前N个当前( to_date = '9999-01-01')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`));

二、代码

窗口函数实现：

select emp_no,salary,sum(salary) over (order by emp_no) as running_total 
from salaries where to_date = '9999-01-01'

复用 salaries 表进行子查询：

主表是a，子查询中复用的表是b，对于小于a.emp_no的b.emp_no所对应的salary进行求和，因为一组a.emp_no对应多个b.emp_no，所以要对a.emp_no进行分组

select a.emp_no,a.salary,sum(b.salary) as running_total 
from salaries a,salaries b
where b.emp_no<=a.emp_no
and a.to_date = '9999-01-01'
and b.to_date = '9999-01-01'
group by a.emp_no

inner join：

把所有小于等于当前编号的表s1和当前编号表s2联立起来，然后按照当前编号分组，计算出所有小于等于
当前标号的工资总数
SELECT s2.emp_no,s2.salary,SUM(s1.salary) AS running_total
FROM salaries AS s1 INNER JOIN salaries AS s2
ON s1.emp_no <= s2.emp_no
WHERE
s1.to_date = "9999-01-01"
AND s2.to_date = "9999-01-01"
GROUP BY s2.emp_no

------------恢复内容结束------------