描述
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));
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 |
示例1：
drop table if exists `salaries` ;
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`));
INSERT INTO salaries VALUES(10001,60117,'1986-06-26','1987-06-26');
INSERT INTO salaries VALUES(10001,62102,'1987-06-26','1988-06-25');
INSERT INTO salaries VALUES(10001,66074,'1988-06-25','1989-06-25');
INSERT INTO salaries VALUES(10001,66596,'1989-06-25','1990-06-25');
INSERT INTO salaries VALUES(10001,66961,'1990-06-25','1991-06-25');
INSERT INTO salaries VALUES(10001,71046,'1991-06-25','1992-06-24');
INSERT INTO salaries VALUES(10001,74333,'1992-06-24','1993-06-24');
INSERT INTO salaries VALUES(10001,75286,'1993-06-24','1994-06-24');
INSERT INTO salaries VALUES(10001,75994,'1994-06-24','1995-06-24');
INSERT INTO salaries VALUES(10001,76884,'1995-06-24','1996-06-23');
INSERT INTO salaries VALUES(10001,80013,'1996-06-23','1997-06-23');
INSERT INTO salaries VALUES(10001,81025,'1997-06-23','1998-06-23');
INSERT INTO salaries VALUES(10001,81097,'1998-06-23','1999-06-23');
INSERT INTO salaries VALUES(10001,84917,'1999-06-23','2000-06-22');
INSERT INTO salaries VALUES(10001,85112,'2000-06-22','2001-06-22');
INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22');
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03');
INSERT INTO salaries VALUES(10002,72527,'1997-08-03','1998-08-03');
INSERT INTO salaries VALUES(10002,72527,'1998-08-03','1999-08-03');
INSERT INTO salaries VALUES(10002,72527,'1999-08-03','2000-08-02');
INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02');
INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');
INSERT INTO salaries VALUES(10003,40006,'1995-12-03','1996-12-02');
INSERT INTO salaries VALUES(10003,43616,'1996-12-02','1997-12-02');
INSERT INTO salaries VALUES(10003,43466,'1997-12-02','1998-12-02');
INSERT INTO salaries VALUES(10003,43636,'1998-12-02','1999-12-02');
INSERT INTO salaries VALUES(10003,43478,'1999-12-02','2000-12-01');
INSERT INTO salaries VALUES(10003,43699,'2000-12-01','2001-12-01');
INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');
输出：
10001|88958|88958
10002|72527|161485
10003|43311|204796
我的解题思路：
本题比较简单，使用窗口函数sum直接求和就行
完整代码：
select emp_no
, salary
, sum(salary) over (order by emp_no) as running_total
from salaries
where to_date = '9999-01-01'
;