题解 | #平均活跃天数和月活人数#

select `month`,
round(sum(来几天)/count(uid),2) avg_active_days,
count(uid) mau
from(
	select date_format(submit_time,"%Y%m") as `month`,uid,	
	count(distinct date(submit_time)) as 来几天
	from exam_record
	where year(submit_time) = 2021 and submit_time is not null
	group by date_format(submit_time,"%Y%m"),uid
	) as 数据过滤
group by `month`;