create table t_access_times(username string,month string,salary int)
row format delimited fields terminated by ',';
load data local inpath '/root/hivedata/t_access_times.dat' into table t_access_times;
A,2015-01,5
A,2015-01,15
B,2015-01,5
A,2015-01,8
B,2015-01,25
A,2015-01,5
A,2015-02,4
A,2015-02,6
B,2015-02,10
B,2015-02,5
1、第一步，先求个用户的月总金额
select username,month,sum(salary) as salary from t_access_times group by username,month
+-----------+----------+---------+--+
| username | month | salary |
+-----------+----------+---------+--+
| A | 2015-01 | 33 |
| A | 2015-02 | 10 |
| B | 2015-01 | 30 |
| B | 2015-02 | 15 |
+-----------+----------+---------+--+
2、第二步，将月总金额表 自己连接 自己连接
select A.*,B.* FROM
(select username,month,sum(salary) as salary from t_access_times group by username,month) A
inner join
(select username,month,sum(salary) as salary from t_access_times group by username,month) B
on
A.username=B.username
where B.month <= A.month
+-------------+----------+-----------+-------------+----------+-----------+--+
| a.username | a.month | a.salary | b.username | b.month | b.salary |
+-------------+----------+-----------+-------------+----------+-----------+--+
| A | 2015-01 | 33 | A | 2015-01 | 33 |
| A | 2015-01 | 33 | A | 2015-02 | 10 |
| A | 2015-02 | 10 | A | 2015-01 | 33 |
| A | 2015-02 | 10 | A | 2015-02 | 10 |
| B | 2015-01 | 30 | B | 2015-01 | 30 |
| B | 2015-01 | 30 | B | 2015-02 | 15 |
| B | 2015-02 | 15 | B | 2015-01 | 30 |
| B | 2015-02 | 15 | B | 2015-02 | 15 |
+-------------+----------+-----------+-------------+----------+-----------+--+
3、第三步，从上一步的结果中
进行分组查询，分组的字段是a.username a.month
求月累计值： 将b.month <= a.month的所有b.salary求和即可
select A.username,A.month,max(A.salary) as salary,sum(B.salary) as accumulate
from
(select username,month,sum(salary) as salary from t_access_times group by username,month) A
inner join
(select username,month,sum(salary) as salary from t_access_times group by username,month) B
on
A.username=B.username
where B.month <= A.month
group by A.username,A.month
order by A.username,A.month;