hive学习02天-访问次数统计

hive的写法和sql类似，却又有一点不一样，本次采用模拟数据编写hql统计访问次数：

求出当月的访问次数，截至当月前的每个月最大访问次数、截至当月前每个用户总的访问次数。

数据表如下

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,2015-03,16
A,2015-03,22
B,2015-03,23
B,2015-03,10
B,2015-03,1

解法一：

--(1)
# 先求出每个用户每个月总访问量
create table record2 as 
select uname,umonth,sum(ucount) as current_month_cnt
from t_access
group by uname,umonth;

# record_2 表中内容为：
A   2015-01     33
A   2015-02     10
A   2015-03     38
B   2015-01     30
B   2015-02     15
B   2015-03     44

--(2)

 

select t1.uname,t1.umonth,t1.current_month_cnt,max(t2.current_month_cnt) as max_cnt,sum(t2.current_month_cnt) as sum_cnt
from record2 t1
join record2 t2
on t1.uname=t2.uname
where t1.umonth >=t2.umonth
group by t1.uname,t1.umonth,t1.current_month_cnt;

# 最终结果：
A   2015-01 33  33  33
A   2015-02 10  43  33
A   2015-03 38  81  38
B   2015-01 30  30  30
B   2015-02 15  45  30
B   2015-03 44  89  44

解法二：

--(1)
# 先求出每个用户每个月总访问量
create table record2 as 
select uname,umonth,sum(ucount) as current_month_cnt
from t_access
group by uname,umonth;

# record_2 表中内容为：
A   2015-01     33
A   2015-02     10
A   2015-03     38
B   2015-01     30 B 2015-02 15 B 2015-03 44

--2

select uname,umonth,current_month_cnt ,
max (current_month_cnt) over(partition by uname order by umonth) as mac_cnt,
sum(current_month_cnt) over(partition by uname order by umonth) as sum_cnt
from
record2;

结果：
A   2015-01 33  33  33
A   2015-02 10  43  33
A   2015-03 38  81  38
B   2015-01 30  30  30
B   2015-02 15  45  30
B   2015-03 44  89  44

 

代码参考：https://www.jianshu.com/p/cdde7125bc77

转载于:https://www.cnblogs.com/students/p/10160376.html