hive累积报表hql

古城微笑少年丶 2022-06-01 11:49 182阅读 0赞

**问题**：下面是一个员工的业绩数据，数据格式是username,date,money,数据如下

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

加载到hive表中了（t\_access\_times）。要求是用sql来计算出，报表name，年月，当月业绩money，这个月和这个月之前的总业绩。(面试的时候还是很有用的)

A,2015-01,33,33
    A,2015-02,10,43
    B,2015-01,30,30
    B,2015-02,15,45

**实现过程**  
1：加载数据到hive

create table t_access_times(username string,month string,salary int) row format delimited fields terminated by ',';
    
    load data local inpath '/home/hadoop/t_access_times.dat' into table t_access_times;

2、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 |  
\+———–+———-+———+–+  
3、第二步，将月总金额表 自己连接 自己连接

select a.username,a.month ,a.salary, b.username, b.month, b.salary 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

\+————-+———-+———–+————-+———-+———–+–+  
| 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 |  
\+————-+———-+———–+————-+———-+———–+–+

4、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  
总的hql为：

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;