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, day(submit_time)))/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 date_format(submit_time,'%Y%m');

难点：1、算月用户活跃总天数。count（distinct uid，day（submit_time)) 即计算每位用户每天的行数。思考：这里可以用sum吗？

易错：括号；函数写法，没写对的时候不会显示颜色。。