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 g...