已有数据表 “t_orgin_weblog”:
+------------------+------------+----------+--+
| col_name | data_type | comment |
+------------------+------------+----------+--+
| valid | string | |
| remote_addr | string | |
| remote_user | string | |
| time_local | string | |
| request | string | |
| status | string | |
| body_bytes_sent | string | |
| http_referer | string | |
| http_user_agent | string | |
+------------------+------------+----------+--+
| true|1.162.203.134| - | 18/Sep/2013:13:47:35| /images/my.jpg | 200| 19939 | "http://www.angularjs.cn/A0d9" | "Mozilla/5.0 (Windows |
| true|1.202.186.37 | - | 18/Sep/2013:15:39:11| /wp-content/uploads/2013/08/windjs.png| 200| 34613 | "http://cnodejs.org/topic/521a30d4bee8d3cb1272ac0f" | "Mozilla/5.0 (Macintosh;|
1、对原始数据进行抽取转换
--将来访url分离出host path query query id
drop table if exists t_etl_referurl;
create table t_etl_referurl as
SELECT a.*,b.*
FROM t_orgin_weblog a LATERAL VIEW parse_url_tuple(regexp_replace(http_referer, "\"", ""), 'HOST', 'PATH','QUERY', 'QUERY:id') b as host, path, query, query_id
2、从前述步骤进一步分离出日期时间形成ETL明细表“t_etl_detail” day tm
drop table if exists t_etl_detail;
create table t_etl_detail as
select b.*,substring(time_local,0,11) as daystr,
substring(time_local,13) as tmstr,
substring(time_local,4,3) as month,
substring(time_local,0,2) as day,
substring(time_local,13,2) as hour
from t_etl_referurl b;
3、对etl数据进行分区(包含所有数据的结构化信息)
drop table t_etl_detail_prt;
create table t_etl_detail_prt(
valid string,
remote_addr string,
remote_user string,
time_local string,
request string,
status string,
body_bytes_sent string,
http_referer string,
http_user_agent string,
host string,
path string,
query string,
query_id string,
daystr string,
tmstr string,
month string,
day string,
hour string)
partitioned by (mm string,dd string);
导入数据
insert into table t_etl_detail_prt partition(mm='Sep',dd='18')
select * from t_etl_detail where daystr='18/Sep/2013';
insert into table t_etl_detail_prt partition(mm='Sep',dd='19')
select * from t_etl_detail where daystr='19/Sep/2013';
分个时间维度统计各referer_host的访问次数并排序
create table t_refer_host_visit_top_tmp as
select referer_host,count(*) as counts,mm,dd,hh from t_display_referer_counts group by hh,dd,mm,referer_host order by hh asc,dd asc,mm asc,counts desc;
4、来源访问次数topn各时间维度URL
取各时间维度的referer_host访问次数topn
select * from (select referer_host,counts,concat(hh,dd),row_number()
over (partition by concat(hh,dd)
order by concat(hh,dd) asc) as od from t_refer_host_visit_top_tmp) t where od<=3;
1、 由于要从大量请求中分辨出用户的各次访问,逻辑相对复杂,通过hive直接实现有困难,因此编写一个mr程序来求出访客访问信息(详见代码)
启动mr程序获取结果:
[hadoop@hdp-node-01 ~]$ hadoop jar weblog.jar cn.itcast.bigdata.hive.mr.UserStayTime /weblog/input /weblog/stayout
2、 将mr的处理结果导入hive表
drop table t_display_access_info_tmp;
create table t_display_access_info_tmp(remote_addr string,firt_req_time string,last_req_time string,stay_long bigint)
row format delimited fields terminated by '\t';
load data inpath '/weblog/stayout4' into table t_display_access_info_tmp;
3、得出访客访问信息表"t_display_access_info"
由于有一些访问记录是单条记录,mr程序处理处的结果给的时长是0,所以考虑给单次请求的停留时间一个默认市场30秒
drop table t_display_access_info;
create table t_display_access_info as
select remote_addr,firt_req_time,last_req_time,
case stay_long
when 0 then 30000
else stay_long
end as stay_long
from t_display_access_info_tmp;
4、统计所有用户停留时间平均值
select avg(stay_long) fromt_display_access_info;
有如下访客访问次数统计表 t_access_times
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 |
…… |
…… |
…… |
需要输出报表:t_access_times_accumulate
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 |
……. |
……. |
……. |
……. |
group by left.month
建表:
create table t_access_times(username string,month string,salary int)load data local inpath '/home/hadoop/t_access_times.dat' into table t_access_times;
检查数据和思路的SQL语句
select username,month,sum(salary) as salary from t_access_times group by username,month
检查数据和思路的SQL语句
A.username=B.username
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;
总结:
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;
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
1、第一步,先求个用户的月总金额
select username,month,sum(salary) as salary from t_access_times group by username,month
+-----------+----------+---------+--+
| username | month | salary |
+-----------+----------+---------+--+
| A | 2015-01 | 33 |
| A | 2015-02 | 10 |
| B | 2015-01 | 30 |
| B | 2015-02 | 15 |
+-----------+----------+---------+--+
2、第二步,将月总金额表 自己连接 自己连接
+-------------+----------+-----------+-------------+----------+-----------+--+
| 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 |
+-------------+----------+-----------+-------------+----------+-----------+--+
max(a.salary不在分组字段,需要使用聚合函数 sum、max) group by (a.username, a.month 两个分组字段)
3、第三步,从上一步的结果中
进行分组查询,分组的字段是a.username a.month
求月累计值: 将b.month <= a.month的所有b.salary求和即可
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//比如右边的必须<=左边的进行求和
group by A.username,A.month
order by A.username,A.month;