HiveSQL——共同使用ip的用户检测问题【自关联问题】

注：参考文章：

SQL 之共同使用ip用户检测问题【自关联问题】-HQL面试题48【拼多多面试题】_hive sql 自关联-CSDN博客文章浏览阅读810次。0 问题描述create table log( uid char(10), ip char(15), time timestamp);insert into log valuesinsert into log values('a', '124', '2019-08-07 12:0:0'),('a', '124', '2019-08-07 13:0:0'),('b', '124', '2019-08-08 12:0:0'),('c', '124', '2019-0._hive sql 自关联https://blog.csdn.net/godlovedaniel/article/details/119858751

0 问题描述

1 数据准备 create table log ( uid string, ip string, login_time string )row format delimited fields terminated by '\t'; insert into log values ('a', '124', '2019-08-07 12:00:00'), ('a', '124', '2019-08-07 13:00:00'), ('b', '124', '2019-08-08 12:00:00'), ('c', '124', '2019-08-09 12:00:00'), ('a', '174', '2019-08-10 12:00:00'), ('b', '174', '2019-08-11 12:00:00'), ('a', '194', '2019-08-12 12:00:00'), ('b', '194', '2019-08-13 13:00:00'), ('c', '174', '2019-08-14 12:00:00'), ('c', '194', '2019-08-15 12:00:00'); 2 数据分析

   共同使用问题，一般此类题型都需要一对多，该问题的解决核心逻辑是自关联。

 完整代码如下：

select t3.uid_1, t3.uid_2 from (select t1.ip, t1.uid as uid_1, t2.uid as uid_2 from (select uid, ip from log group by uid, ip) t1 join (select uid, ip from log group by uid, ip) t2 where t1.ip = t2.ip and t1.uid < t2.uid) t3 group by t3.uid_1, t3.uid_2 having count(ip) >= 3;

代码分析：

step1: 获取自关联的结果集

select t1.ip, t1.uid as uid_1, t2.uid as uid_2 from (select uid, ip from log group by uid, ip) t1 join(select uid, ip from log group by uid, ip) t2 on t1.ip = t2.ip;

step2: 由于数据会两两出现，所以a,b和 b,a实际上是一样的，需要过滤掉这部分重复数据，只需要选出 t1.uid < t2.uid，即过滤掉a,b这组数据。hive中不支持不等连接，故使用where语句

select t1.ip, t1.uid as uid_1, t2.uid as uid_2 from (select uid, ip from log group by uid, ip) t1 join (select uid, ip from log group by uid, ip) t2 where t1.ip = t2.ip and t1.uid < t2.uid;

step3:按照组合键分组，并过滤出符合条件的用户

select t3.uid_1, t3.uid_2 from (select t1.ip, t1.uid as uid_1, t2.uid as uid_2 from (select uid, ip from log group by uid, ip) t1 join (select uid, ip from log group by uid, ip) t2 where t1.ip = t2.ip and t1.uid < t2.uid) t3 group by t3.uid_1, t3.uid_2 having count(ip) >= 3; 3 小结

    本案例题型属于：“共同xx”，例如：共同好友、互相认识、共同使用等。遇到这类关键字的时候，往往可以采用自关联的方式解决。（笛卡尔积：一对多；去重取一）