题解 | #2021年11月每天的人均浏览文章时长#

select 
    dt,round(sum(sumtime)/count(cnt),1) as avg_viiew_len_sec
from (select 
    dt,uid,count(uid) as cnt,
    sum(times) as sumtime
from (select
            uid,
            artical_id,
            timestampdiff(second, in_time, out_time) as times,
            date (in_time) as dt
        from
            tb_user_log where artical_id!= 0 and  substring(in_time,1,7) = '2021-11') t 
    group by dt,uid) p 
    group by dt order by avg_viiew_len_sec ;

一次答不对 每次都是错了一步步改 心累