SQL查询语句求出用户的连续登陆天数

时间:2022-11-10 08:21:48

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
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
-- 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

?
1
2
3
4
5
6
7
8
9
10
11
12
13
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

SQL查询语句求出用户的连续登陆天数

以上就是SQL查询语句求出用户的连续登陆天数的详细内容,更多关于SQL语句求用户的连续登陆天数的资料请关注服务器之家其它相关文章!

原文链接:https://blog.csdn.net/Heng_bigdate_yan/article/details/120643783