题解 | #平均活跃天数和月活人数#

select distinct(concat(year(start_time),'0',month(start_time))) as month,
round(count(distinct uid ,day(submit_time))/count(distinct uid),2) as avg_active_days,
count(distinct uid ) as  mau

from exam_record
where year(start_time)=2021 and submit_time is not null
group by month