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,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
group by DATE_FORMAT(submit_time,'%Y%m')

又一次做到这道题，除了知识的综合使用。

这道题比较重要的就是 时间函数的应用，尤其是date_format的使用。