题解 | #平均活跃天数和月活人数#
平均活跃天数和月活人数
http://www.nowcoder.com/practice/9e2fb674b58b4f60ac765b7a37dde1b9
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
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