背景web
对于数据开发人员来讲,手写sql是比较熟悉的了,就有这样一道题,面试时须要手写sql,这就是很是经典的连续登陆问题,大厂小厂都爱问,这种题说简单也不简单,说难也不难,关键是要有思路。
面试
真题sql
hql统计连续登录的三天及以上的用户ruby
这个问题能够扩展到不少类似的问题:连续几个月充值会员、连续天数有商品卖出、连续打车、连续逾期。bash
数据提供
微信
用户ID、登入日期 user01,2018-02-28 user01,2018-03-01 user01,2018-03-02 user01,2018-03-04 user01,2018-03-05 user01,2018-03-06 user01,2018-03-07 user02,2018-03-01 user02,2018-03-02 user02,2018-03-03 user02,2018-03-06
输出字段
app
+---------+--------+-------------+-------------+--+| uid | times | start_date | end_date |+---------+--------+-------------+-------------+--+
群内讨论编辑器
这道题在群里发出后,你们就展开了激烈的讨论:学习
解决方案大数据
能够看出来,有不少种不一样的解决方案。
这里就为你们提供一种比较常见的方案:
建表
create table wedw_dw.t_login_info( user_id string COMMENT '用户ID',login_date date COMMENT '登陆日期')row format delimitedfields terminated by ',';
导数据
hdfs dfs -put /test/login.txt /data/hive/test/wedw/dw/t_login_info/
验证数据
select * from wedw_dw.t_login_info;+----------+-------------+--+| user_id | login_date |+----------+-------------+--+| user01 | 2018-02-28 || user01 | 2018-03-01 || user01 | 2018-03-02 || user01 | 2018-03-04 || user01 | 2018-03-05 || user01 | 2018-03-06 || user01 | 2018-03-07 || user02 | 2018-03-01 || user02 | 2018-03-02 || user02 | 2018-03-03 || user02 | 2018-03-06 |+----------+-------------+--+
解决方案
select t2.user_id as user_id,count(1) as times,min(t2.login_date) as start_date,max(t2.login_date) as end_datefrom( select t1.user_id ,t1.login_date ,date_sub(t1.login_date,rn) as date_diff from ( select user_id ,login_date ,row_number() over(partition by user_id order by login_date asc) as rn from wedw_dw.t_login_info ) t1) t2group by t2.user_id,t2.date_diffhaving times >= 3;
结果
+----------+--------+-------------+-------------+--+| user_id | times | start_date | end_date |+----------+--------+-------------+-------------+--+| user01 | 3 | 2018-02-28 | 2018-03-02 || user01 | 4 | 2018-03-04 | 2018-03-07 || user02 | 3 | 2018-03-01 | 2018-03-03 |+----------+--------+-------------+-------------+--+
思路
先把数据按照用户id分组,根据登陆日期排序
select user_id ,login_date ,row_number() over(partition by user_id order by login_date asc) as rn from wedw_dw.t_login_info
+----------+-------------+-----+--+| user_id | login_date | rn |+----------+-------------+-----+--+| user01 | 2018-02-28 | 1 || user01 | 2018-03-01 | 2 || user01 | 2018-03-02 | 3 || user01 | 2018-03-04 | 4 || user01 | 2018-03-05 | 5 || user01 | 2018-03-06 | 6 || user01 | 2018-03-07 | 7 || user02 | 2018-03-01 | 1 || user02 | 2018-03-02 | 2 || user02 | 2018-03-03 | 3 || user02 | 2018-03-06 | 4 |+----------+-------------+-----+--+
2.用登陆日期减去排序数字rn,获得的差值日期若是是相等的,则说明这两天确定是连续的
select t1.user_id ,t1.login_date ,date_sub(t1.login_date,rn) as date_diff from ( select user_id ,login_date ,row_number() over(partition by user_id order by login_date asc) as rn from wedw_dw.t_login_info ) t1 ;
+----------+-------------+-------------+--+| user_id | login_date | date_diff |+----------+-------------+-------------+--+| user01 | 2018-02-28 | 2018-02-27 || user01 | 2018-03-01 | 2018-02-27 || user01 | 2018-03-02 | 2018-02-27 || user01 | 2018-03-04 | 2018-02-28 || user01 | 2018-03-05 | 2018-02-28 || user01 | 2018-03-06 | 2018-02-28 || user01 | 2018-03-07 | 2018-02-28 || user02 | 2018-03-01 | 2018-02-28 || user02 | 2018-03-02 | 2018-02-28 || user02 | 2018-03-03 | 2018-02-28 || user02 | 2018-03-06 | 2018-03-02 |+----------+-------------+-------------+--+
3.根据user_id和日期差date_diff 分组,最小登陆日期即为这次连续登陆的开始日期start_date,最大登陆日期即为结束日期end_date,登陆次数即为分组后的count(1)
select t2.user_id as user_id,count(1) as times,min(t2.login_date) as start_date,max(t2.login_date) as end_datefrom( select t1.user_id ,t1.login_date ,date_sub(t1.login_date,rn) as date_diff from ( select user_id ,login_date ,row_number() over(partition by user_id order by login_date asc) as rn from wedw_dw.t_login_info ) t1) t2group by t2.user_id,t2.date_diffhaving times >= 3;
+----------+--------+-------------+-------------+--+| user_id | times | start_date | end_date |+----------+--------+-------------+-------------+--+| user01 | 3 | 2018-02-28 | 2018-03-02 || user01 | 4 | 2018-03-04 | 2018-03-07 || user02 | 3 | 2018-03-01 | 2018-03-03 |+----------+--------+-------------+-------------+--+
结束语
以上仅提供了一种解决方案,小伙伴有其余方案的话,能够进群交流哦
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