题解 | #

select zb1.author_id,author_level,days_cnt
from 
(select author_id,count(fuzhu) days_cnt /*通过连登录天数*/
from
(select distinct answer_date,author_id,
answer_date-row_number() over(partition by author_id order by answer_date) fuzhu /*建立一个差值辅助列,值相等次数即为连续登录天数*/
from answer_tb) zb
group by author_id,fuzhu
having days_cnt >=3) zb1 join author_tb at on zb1.author_id=at.author_id
order by zb1.author_id