Hive基本操作(二)——Hive实战案例-级联求和
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Hive实战案例——级联求和
源数据:
访客 月份 访问次数
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
--------------------------------
操作结果Result:
访客 月份 月访问总计 累计访问总计
A 2015-01 33 33
A 2015-02 10 43
B 2015-01 30 30
B 2015-02 15 45
================================
username month salary
===================================
create table if not exists t_access_times(
username string,month string,salary int
)row format delimited fields terminated by " ";
=============================================
第一步:
select username,month,sum(salary) as salary from t_access_times group by username,month;
A 2015-01 33
A 2015-02 10
B 2015-01 30
B 2015-02 15
---------------------------------------------
A 2015-01 33
A 2015-02 10
B 2015-01 30
B 2015-02 15
=============================================
第二步:
select * 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 2015-01 33 A 2015-01 33
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 2015-02 15 B 2015-01 30
B 2015-02 15 B 2015-02 15
===============================================
第三步:
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;
Result:
A 2015-01 33 33
A 2015-02 10 43
B 2015-01 30 30
B 2015-02 15 45
源数据:
访客 月份 访问次数
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
--------------------------------
操作结果Result:
访客 月份 月访问总计 累计访问总计
A 2015-01 33 33
A 2015-02 10 43
B 2015-01 30 30
B 2015-02 15 45
================================
username month salary
===================================
create table if not exists t_access_times(
username string,month string,salary int
)row format delimited fields terminated by " ";
=============================================
第一步:
select username,month,sum(salary) as salary from t_access_times group by username,month;
A 2015-01 33
A 2015-02 10
B 2015-01 30
B 2015-02 15
---------------------------------------------
A 2015-01 33
A 2015-02 10
B 2015-01 30
B 2015-02 15
=============================================
第二步:
select * 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 2015-01 33 A 2015-01 33
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 2015-02 15 B 2015-01 30
B 2015-02 15 B 2015-02 15
===============================================
第三步:
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;
Result:
A 2015-01 33 33
A 2015-02 10 43
B 2015-01 30 30
B 2015-02 15 45
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