题解 | #查询连续登陆的用户#

with tmp as (
    select *, row_number() over(partition by user_id order by log_time asc) as r from login_tb join register_tb using(user_id)
)

select user_id from (select user_id, date_sub(date(log_time), interval r DAY) as daygap from tmp) as t
group by user_id, daygap
having count(daygap)>= 3
order by user_id asc

模板题目，窗口函数会按照order by 排序，daygap是个日期，此外新学到了 a join b using()