解決sql實現累計報表問題。
需求
有如下訪客訪問次數統計表 t_access_times
|
訪客 |
月份 |
訪問次數 |
|
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 |
|
…… |
…… |
…… |
需要輸出報表：t_access_times_accumulate
|
訪客 |
月份 |
月訪問總計 |
累計訪問總計 |
|
A |
2015-01 |
33 |
33 |
|
A |
2015-02 |
10 |
43 |
|
……. |
……. |
……. |
……. |
|
B |
2015-01 |
30 |
30 |
|
B |
2015-02 |
15 |
45 |
|
……. |
……. |
……. |
……. |
如果是Hive則先創建表
create table t_access_times(username string,month string,salary int)
row format delimited fields terminated by ',';
將數據加載到表
load data local inpath '/home/hadoop/t_access_times.dat' into table t_access_times;
表數據
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，先求個用戶的月總金額
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，將月總金額表 自己連接 自己連接 實際上是利用笛卡爾積
+-------------+----------+-----------+-------------+----------+-----------+--+
| 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，從上一步的結果中
進行分組查詢，分組的字段是a.username a.month
求月累計值： 將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;