SQL查询语句求出用户的连续登陆天数
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2022-06-17 19:06:04
一、题目描述求解用户登陆信息表中,每个用户连续登陆平台的天数,连续登陆基础为汇总日期必须登陆,表中每天只有一条用户登陆数据(计算中不涉及天内去重)。表描述:user_id:用户的id; s...
一、题目描述
求解用户登陆信息表中,每个用户连续登陆平台的天数,连续登陆基础为汇总日期必须登陆,表中每天只有一条用户登陆数据(计算中不涉及天内去重)。
表描述:user_id:用户的id;
sigin_date:用户的登陆日期。
二、解法分析
注:求解过程有多种方式,下述求解解法为笔者思路,其他解法可在评论区交流。
思路:
该问题的突破的在于登陆时间,计算得到连续登陆标识,以标识分组为过滤条件,得到连续登陆的天数,最后以user_id分组,以count()函数求和得到每个用户的连续登陆天数。
连续登陆标识 =(当日登陆日期 - 用户的登陆日期)- 开窗排序的顺序号(倒序)
三、求解过程及结果展示
1.数据准备
-- 1.建表语句 drop table if exists test_sigindate_cnt; create table test_sigindate_cnt( user_id string ,sigin_date string ) ; -- 2.测试数据插入语句 insert overwrite table test_sigindate_cnt select 'uid_1' as user_id,'2021-08-03' as sigin_date union all select 'uid_1' as user_id,'2021-08-04' as sigin_date union all select 'uid_1' as user_id,'2021-08-01' as sigin_date union all select 'uid_1' as user_id,'2021-08-02' as sigin_date union all select 'uid_1' as user_id,'2021-08-05' as sigin_date union all select 'uid_1' as user_id,'2021-08-06' as sigin_date union all select 'uid_2' as user_id,'2021-08-01' as sigin_date union all select 'uid_2' as user_id,'2021-08-05' as sigin_date union all select 'uid_2' as user_id,'2021-08-02' as sigin_date union all select 'uid_2' as user_id,'2021-08-06' as sigin_date union all select 'uid_3' as user_id,'2021-08-04' as sigin_date union all select 'uid_3' as user_id,'2021-08-06' as sigin_date union all select 'uid_4' as user_id,'2021-08-03' as sigin_date union all select 'uid_4' as user_id,'2021-08-02' as sigin_date ;
2.计算过程
select user_id ,count(1) as sigin_cnt from ( select user_id ,datediff('2021-08-06',sigin_date) as data_diff ,row_number() over (partition by user_id order by sigin_date desc) as row_num from test_sigindate_cnt ) t where data_diff - row_num = -1 group by user_id ;
3.计算结果及预期结果对比
3.1 预期结果
汇总日期 | 用户id | 登陆天数 |
2021-08-06 | uid_1 | 6 |
2021-08-06 | uid_2 | 2 |
2021-08-06 | uid_3 | 1 |
3.2 计算结果
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