题解 | #2021年11月每天的人均浏览文章时长#

select dt,
       round(sum(sec)/count(distinct uid),1) avg_view_len_sec
from (select uid,
             date(in_time) dt, 
             TIMESTAMPDIFF(SECOND,in_time,out_time) sec
      from tb_user_log
      where date_format(in_time,'%Y-%m') = '2021-11' 
            and artical_id != 0) a
group by dt
order by avg_view_len_sec