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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 l…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
selectt3.uid_1, t3.uid_2
from (selectt1.ip,t1.uid as uid_1,t2.uid as uid_2from (select uid, ip from log group by uid, ip) t1join(select uid, ip from log group by uid, ip) t2where t1.ip t2.ipand 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
selectt1.ip,t1.uid as uid_1,t2.uid as uid_2
from (select uid, ip from log group by uid, ip) t1join(select uid, ip from log group by uid, ip) t2on 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
selectt1.ip,t1.uid as uid_1,t2.uid as uid_2
from (select uid, ip from log group by uid, ip) t1join (select uid, ip from log group by uid, ip) t2where 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
selectt3.uid_1, t3.uid_2
from (selectt1.ip,t1.uid as uid_1,t2.uid as uid_2from (select uid, ip from log group by uid, ip) t1join(select uid, ip from log group by uid, ip) t2where t1.ip t2.ipand 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
