题解 | 查询连续登陆的用户

with by_day as (
select user_id,log_time,DATE_SUB(date(log_time), INTERVAL (row_number() over(partition by user_id order by log_time)) DAY) day from login_tb
where user_id in (select user_id from register_tb))
select user_id from by_day group by  user_id,day
having count(*)>=3 order by user_id

row_number() over(partition by user_id order by log_time) rn 给日期表个序号

DATE_SUB(date(log_time), INTERVAL rn DAY) 原日期减去序号 如果是按照顺序来的，那么就是相同的一天

select user_id from by_day group by user_id,day 按照人和天来分组，相同天数大于等于三的 就是连续登录大于等于三天的