Sql经典面试题目------累计报表

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;

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;