Hive级联求和

Posted 2022-06-06 YaoYong_BigData

create table t_salary_detail(username string,month string,salary int)
row format delimited fields terminated by ',';

load data local inpath '/root/hivedata/t_salary_detail.dat' into table t_salary_detail;

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
A,2015-03,7
A,2015-03,9
B,2015-03,11
B,2015-03,6

需求：统计每个用户每个月获得多少小费？
分组维度：  username  month

1、第一步，先求每个用户的月总金额
select username,month,sum(salary) as salary from t_salary_detail group by username,month;

+-----------+----------+---------+-
| username  |  month   | salary  |   
+-----------+----------+---------+-
| A         | 2015-01  | 33      |    
| A         | 2015-02  | 10      |    
| A         | 2015-03  | 16      |    
| B         | 2015-01  | 33      |    
| B         | 2015-02  | 15      |    
| B         | 2015-03  | 17      |    
+-----------+----------+---------+--+

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) as salary from t_salary_detail group by username,month) A 
inner join 
(select username,month,sum(salary) as salary from t_salary_detail 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           | 2015-02  | 10        | A           | 2015-01  | 33        |
| A           | 2015-02  | 10        | A           | 2015-02  | 10        |
| A           | 2015-03  | 16        | A           | 2015-01  | 33        |
| A           | 2015-03  | 16        | A           | 2015-02  | 10        |
| A           | 2015-03  | 16        | A           | 2015-03  | 16        |
| B           | 2015-01  | 30        | B           | 2015-01  | 30        |
| B           | 2015-02  | 15        | B           | 2015-01  | 30        |
| B           | 2015-02  | 15        | B           | 2015-02  | 15        |
| B           | 2015-03  | 17        | B           | 2015-01  | 30        |
| B           | 2015-03  | 17        | B           | 2015-02  | 15        |
| B           | 2015-03  | 17        | B           | 2015-03  | 17        |
+-------------+----------+-----------+-------------+----------+-----------+--+

不加where B.month <= A.month的执行结果：

​​​​​​​

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_salary_detail group by username,month) A 
inner join 
(select username,month,sum(salary) as salary from t_salary_detail 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;

 

+-------------+----------+---------+-------------+--+
| a.username  | a.month  | salary  | accumulate  |
+-------------+----------+---------+-------------+--+
| A           | 2015-01  | 33      | 33          |
| A           | 2015-02  | 10      | 43          |
| A           | 2015-03  | 16      | 59          |
| B           | 2015-01  | 30      | 30          |
| B           | 2015-02  | 15      | 45          |
| B           | 2015-03  | 17      | 62          |
+-------------+----------+---------+-------------+--+