题解 | #平均活跃天数和月活人数#

之前不知道可以count（distinct 字段1，字段2）达到下边代码中子查询的目的。所以比较累赘。

select month,
round(count(day)/count(distinct uid),2)avg_active_days
,count(distinct uid) mau from (
select  distinct uid, 
date_format(start_time,"%Y-%m-%d") day  ,
date_format(start_time,"%Y%m") month  from exam_record
where score is not null and start_time like '2021%'
)a
group by month