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SQL查询语句求出用户的连续登陆天数_MsSql_

2023-05-26 376人已围观

简介 SQL查询语句求出用户的连续登陆天数_MsSql_

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 ;