1、题目描述
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`));
输出格式:
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`));
输出格式:
| emp_no | salary | running_total |
|---|---|---|
| 10001 | 88958 | 88958 |
| 10002 | 72527 | 161485 |
| 10003 | 43311 | 204796 |
| 10004 | 74057 | 278853 |
| 10005 | 94692 | 373545 |
| 10006 | 43311 | 416856 |
| 10007 | 88070 | 504926 |
| 10009 | 95409 | 600335 |
| 10010 | 94409 | 694744 |
| 10011 | 25828 | 720572 |
2、代码
SELECT s1.emp_no, s1.salary, (SELECT SUM(s2.salary) FROM salaries AS s2 WHERE s2.emp_no <= s1.emp_no AND s2.to_date = '9999-01-01') AS running_total FROM salaries AS s1 WHERE s1.to_date = '9999-01-01' ORDER BY s1.emp_no；