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SQL查询语句求出用户的连续登陆天数代码示例

时间:2021-10-11 17:36:45 编辑:袖梨 来源:转载

本篇文章小编给大家分享一下SQL查询语句求出用户的连续登陆天数代码示例,文章代码介绍的很详细,小编觉得挺不错的,现在分享给大家供大家参考,有需要的小伙伴们可以来看看。

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

SQL查询语句求出用户的连续登陆天数代码示例

3.2 计算结果

SQL查询语句求出用户的连续登陆天数代码示例

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