题解 | 平均活跃天数和月活人数

select month,round(sum(num)/count(uid),2) as avg_active_days,count(uid) as mau
from(
select month,uid,count(distinct date(start_time)) num
from(
select *,date_format(date(start_time),'%Y%m') as month
from exam_record
where score != '(NULL)'
and left(start_time,4) = '2021') as t1
group by month,uid) as t2
group by month

先计算每月每个用户的活跃次数

再计算别的