# 以author_id分组，将回答天数>=3天的删选出来，之后再看是否连续（将日期放到一个字符串中？）
select a.author_id, author_level, answer_days
from (
    select count(distinct answer_date) as answer_days, author_id
    from answer_tb
    group by author_id
    having answer_days >= 3
) as a
left join author_tb as b
on a.author_id = b.author_id
order by author_id;