题解 | #平均活跃天数和月活人数#

select
    month
    ,round(sum(active_days)/count(uid),2) as avg_active_days
    ,count(uid) as mau
from
(
    select 
        date_format(submit_time,'%Y%m') as month
        ,uid
        ,count(distinct day(submit_time)) as active_days
    from exam_record
    where year(submit_time) in (2021)
    group by 
        date_format(submit_time,'%Y%m')
        ,uid
) t
group by
    month