题解有个不严谨的地方就是，如果用户是近7天内的新用户，并且在7天内不止活跃1次，用这个方法就没办法确定用户是“忠实”还是“新晋”,想在原题解的基础上做一下修改，确保题解的严谨性，没想到意外发现times根本就不用去计算，真是没想到能在确保题解严谨性的同时进一步精简代码，具体如下： select grade ,round(count(*)/(select count(distinct uid) from tb_user_log),2) as ration from (select case when min_gap<7 and max_map>=7 then "忠实用户" when min_gap<7 and max_map<7 then "新晋用户" when min_gap>=7 and min_gap<30 then "沉睡用户" when min_gap>=30 then "流失用户" end as grade, uid from(select uid,min(gap) as min_gap,max(gap) as max_map from(select uid,timestampdiff(day,in_time,(select max(in_time) from tb_user_log)) as gap from tb_user_log) tmp group by uid) base) info group by grade order by ration desc,grade