SQL查询语句求出用户的连续登陆天数

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 

3.2 计算结果

来源：https://blog.csdn.net/Heng_bigdate_yan/article/details/120643783