hive的寫法和sql類似,卻又有一點不一樣,本次采用模擬數據編寫hql統計訪問次數:
求出當月的訪問次數,截至當月前的每個月最大訪問次數、截至當月前每個用戶總的訪問次數。
數據表如下
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) # 先求出每個用戶每個月總訪問量 create table record2 as
select uname,umonth,sum(ucount) as current_month_cnt
from t_access
group by uname,umonth; # record_2 表中內容為: 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;
# 最終結果: 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) # 先求出每個用戶每個月總訪問量 create table record2 as
select uname,umonth,sum(ucount) as current_month_cnt
from t_access
group by uname,umonth; # record_2 表中內容為: 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;
結果: 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
代碼參考:https://www.jianshu.com/p/cdde7125bc77