题解 | 查询连续登陆的用户

  select
    lt.user_id
from login_tb lt
inner join register_tb rt on lt.user_id=rt.user_id
inner join login_tb lt1 on lt.user_id=lt1.user_id and DATE(date_add(lt.log_time,interval 1 day)) = DATE(lt1.log_time)
inner join login_tb lt2 on lt.user_id=lt2.user_id and DATE(date_add(lt.log_time,interval 2 day)) = DATE(lt2.log_time)
order by lt.user_id asc