题解 | #平均活跃天数和月活人数#

select submit_month as month, round(count(distinct uid,submit_day)/count(distinct uid),2) as avg_active_days,count(distinct uid) as mau
from (
select id,uid,exam_id,submit_time,date_format(submit_time,'%Y%m%d') as submit_day,date_format(submit_time,'%Y%m')as submit_month
from exam_record
where submit_time is not null and year(submit_time)=2021
) t2
group by month;