根据访问次数统计表，得到累计访问总计
- 建表，load数据
t_access_times.dat
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
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;
+--------------------------+-----------------------+------------------------+--+
| t_access_times.username | t_access_times.month | t_access_times.salary |
+--------------------------+-----------------------+------------------------+--+
| 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 |
+--------------------------+-----------------------+------------------------+--
- 求每个用户的月总金额
select username,month,sum(salary) from t_access_times group by username,month;
+-----------+----------+------+--+
| username | month | _c2 |
+-----------+----------+------+--+
| A | 2015-01 | 33 |
| A | 2015-02 | 10 |
| B | 2015-01 | 30 |
| B | 2015-02 | 15 |
+-----------+----------+------+--+
- 把表自己inner join
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
+-------------+----------+-----------+-------------+----------+-----------+--+
| 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 |
+-------------+----------+-----------+-------------+----------+-----------+--+
- 生成累计值
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;
+-------------+----------+---------+-------------+--+
| a.username | a.month | salary | accumulate |
+-------------+----------+---------+-------------+--+
| A | 2015-01 | 33 | 33 |
| A | 2015-02 | 10 | 43 |
| B | 2015-01 | 30 | 30 |
| B | 2015-02 | 15 | 45 |
+-------------+----------+---------+-------------+
分组查询求月累计值。
为什么要max（salary）?
salary不是分组字段，只能由聚合函数求得，不然不知道选哪个。sum avg max
最后order by 使全局有序，原始数据无序，最后有可能无序。
