SQL查询语句求出用户的连续登陆天数 SQL查询语句求出用户的连续登

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；
【SQL查询语句求出用户的连续登陆天数】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_dateunion allselect 'uid_1' as user_id,'2021-08-04' as sigin_date union allselect 'uid_1' as user_id,'2021-08-01' as sigin_dateunion allselect 'uid_1' as user_id,'2021-08-02' as sigin_dateunion allselect 'uid_1' as user_id,'2021-08-05' as sigin_dateunion allselect 'uid_1' as user_id,'2021-08-06' as sigin_dateunion allselect 'uid_2' as user_id,'2021-08-01' as sigin_dateunion allselect 'uid_2' as user_id,'2021-08-05' as sigin_dateunion allselect 'uid_2' as user_id,'2021-08-02' as sigin_dateunion allselect 'uid_2' as user_id,'2021-08-06' as sigin_dateunion allselect 'uid_3' as user_id,'2021-08-04' as sigin_dateunion allselect 'uid_3' as user_id,'2021-08-06' as sigin_dateunion allselect 'uid_4' as user_id,'2021-08-03' as sigin_dateunion allselect 'uid_4' as user_id,'2021-08-02' as sigin_date;

2.计算过程

selectuser_id,count(1) as sigin_cntfrom(selectuser_id,datediff('2021-08-06',sigin_date)as data_diff,row_number() over (partition by user_id order by sigin_date desc) as row_numfromtest_sigindate_cnt) twheredata_diff - row_num = -1group by user_id;

3.计算结果及预期结果对比
3.1 预期结果