题解 | 就这样吧，可以简化很多，犯懒了，不简化了，作差方式

select
t1.author_id,
ab.author_level,
t1.pt
from(
select
t.author_id,
count(*) as pt
from(
select
author_id,
answer_date,
row_number()over(partition by author_id order by answer_date) as m
from answer_tb
group by author_id,answer_date
having count(distinct issue_id)>=1
) as t 
group by subdate(t.answer_date,t.m),t.author_id
having count(*)>=3
) as t1 left join author_tb ab on t1.author_id=ab.author_id
order by t1.author_id asc