牛客——SQL254 统计salary的累计和running_total

发布时间 2023-09-07 08:57:35作者: 莫离y

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'
;