mysql统计累计和

题目：

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`));

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

 

方法一：窗口函数

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

 

方法二：自连接

形式1：

select a.emp_no, a.salary, sum(b.salary) as running_total
from salaries a inner join salaries b
on b.emp_no <= a.emp_no
where a.to_date = '9999-01-01'
and b.to_date = '9999-01-01'
group by a.emp_no,a.salary
#注意mysql必须加上a.salary,sqlite可以不加

 

形式2：

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

 

形式3：

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

 

顺便学习在workbench创建表，插入数据：

https://www.bilibili.com/video/BV18y4y1H7RD?spm_id_from=333.337.search-card.all.click

 

create database test;
Use test;
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');

使用上述方法，结果如下：