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 year (submit_time) = 2021 and submit_time is not null group by month; 用户平均月活跃天 = (用户月活...