Hive面试题——累计求和

需求：

有如下访客访问次数统计表 t_access_times

访客	月份	访问次数
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
……	……	……

 

需要输出报表：t_access_times_accumulate

访客	月份	月访问总计	累计访问总计
A	2015-01	33	33
A	2015-02	10	43
…….	…….	…….	…….
B	2015-01	30	30
B	2015-02	15	45
…….	…….	…….	…….

思路：

1、第一步，先求个用户的月总金额

select username,month,sum(salary) 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

select A.*,B.*
from 
(select username,month,sum(salary) salary from t_access_times group by username,month) A 
join 
(select username,month,sum(salary) salary from t_access_times group by username,month) B
on
A.username=B.username;

+-------------+----------+-----------+-------------+----------+-----------+--+
| 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) salary,sum(B.salary) accumulate
from 
(select username,month,sum(salary) salary from t_access_times group by username,month) A 
join 
(select username,month,sum(salary) 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;