题解 | #平均活跃天数和月活人数#

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