在大資料面試中，Hive知識的考察大部分會問級聯求和，業務場景雖然有很多種，比如說，年收入，月收入統計；訪客訪問次數年統計，月統計。等等。但是基本根源知識是級聯求和，本文就以訪客訪問統計為例。

1、基本需求

根據訪客的每日訪問資訊，進行累計訪問：

• 輸入資料：

有如下訪客訪問次數統計表 t_access_times

為了減輕計算複雜度，去掉了天的資訊只留下了年月。

• 輸出資料：

需要輸出報表：t_access_times_accumulate

2、實現步驟

2.1 原始資料

vi /home/chunsoft/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

2.2 建立表

jdbc:hive2://172.16.208.128:10000>create table t_access_times(username string,month string,access_time int)
row format delimited fields terminated by ',';

2.3 將資料載入到表中

jdbc:hive2://172.16.208.128:10000>
load
 
 data local inpath '/home/chunsoft/t_access_times.dat' into table t_access_times;

2.4 計算單個使用者的月訪問次數

jdbc:hive2://172.16.208.128:10000>
select username,month,sum(access_time) as access_time from t_access_times group by username,month;

+-----------+----------+---------+--+
| username  |  month   | access_time  |
+-----------+----------+---------+--+
 

| A         | 2015-01  | 33      |
| A         | 2015-02  | 10      |
| B         | 2015-01  | 30      |
| B         | 2015-02  | 15      |
+-----------+----------+---------+--+

2.4 上面的表自己跟自己進行inner join方便求總訪問

select A.*,B.* from
(select username,month,sum(access_time) as access_time from t_access_times group by username,month) A
inner join
(select username,month,sum(access_time) as access_time from t_access_times group by username,month) B
on
A.username=B.username

+-------------+----------+-----------+-------------+----------+-----------+--+
| a.username  | a.month  | a.access_time  | b.username  | b.month  | b.access_time  |
+-------------+----------+-----------+-------------+----------+-----------+--+
| 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        |
+-------------+----------+-----------+-------------+----------+-----------+--+

2.5 根據當前月份的大小來判斷累計求和並排序

select A.username,A.month,max(A.access_time) access_time,sum(B.access_time) accumulate
from
  (select username,month,sum(access_time) as access_time from t_access_times group by username,month) A
  inner join
  (select username,month,sum(access_time) as access_time 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  | access_time  | accumulate  |
+-------------+----------+---------+-------------+--+
| A           | 2015-01  | 33      | 33          |
| A           | 2015-02  | 10      | 43          |
| B           | 2015-01  | 30      | 30          |
| B           | 2015-02  | 15      | 45          |

這裡面的max(A.access_time) 起到一個聚合作用，保留一條結果。