这道题做了好久老是报错,后面才发现是理解错了【用户平均月活跃天数】我以为是用户平均月活跃天数=count(uid)/count(distinct uid)但后面发现,一个用户在[一天内]可能会提交[多份]试卷,所以需要按照uid和天数来去重,所以:用户平均月活跃天数=count(distinct uid,date_format(submit_time,'%Y%m%d')/distinct uid 代码如下: select date_format(submit_time,'%Y%m') as month,round(count(distinct uid,date_format(submit_t...