sql连续登录问题

今天刷到一个视频，需求是计算一个库表里，查询2022年1月份里连续三天登录的用户以及连续登录的时间段。
首先，t_login记录了用户每次登录的时间信息。
一天有多次登录行为，只需记录一条;
(使用函数导致索引失效)

select distinct uid,date(login_time) ymdfrom t_loginwhere login_time between '2022-01-01 00:00:00' and timestamp '2022-01-31 23:59:59';

比较粗暴方式，将上面的表自关联三次，缺点，扩展性差。

select t1.uid,t1.ymd,t2.ymd,t3.ymd from(select distinct uid,date(login_time) ymdfrom t_loginwhere login_time between '2022-01-01 00:00:00' and timestamp '2022-01-31 23:59:59') t1join(select distinct uid,date(login_time) ymdfrom t_loginwhere login_time between '2022-01-01 00:00:00' and timestamp '2022-01-31 23:59:59') t2on(t1.uid=t2.uid and datediff(t2.ymd,t1.tmd)=1)join(select distinct uid,date(login_time) ymdfrom t_loginwhere login_time between '2022-01-01 00:00:00' and timestamp '2022-01-31 23:59:59') t3on(t2.uid=t3.uid and datediff(t3.ymd,t2.tmd)=1);

方式二

使用窗口函数，分区限制条件，得到

with t1 as (select distinct uid,date(login_time) ymd            from t_login            where login_time between timestamp '2022-01-01 00:00:00'                             and timestamp '2022-01-31 23:59:59')                        select uid,ymd,row_number() over (partition by uid order by ymd) numfrom t1;          

只要不连续，插值一定存在结果跳跃。

with t1 as (select distinct uid,date(login_time) ymd            from t_login            where login_time between timestamp '2022-01-01 00:00:00'                             and timestamp '2022-01-31 23:59:59')  t2 as(select uid,ymd,date_sub(ymd,interval row_number() over (partition by uid order by ymd) day) sub_datefrom t1)select uid,min(ymd),max(ymd),count(*)from t2group by uid,sub_datehaving count(*)>=3;

妈的后悔了，浪费时间了。哎没时间感叹。
作者：为什么不好好卖蛋饼

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