题解 | #2021年11月每天的人均浏览文章时长#

select
date_format(in_time,'%Y-%m-%d') dt
,round(sum(timestampdiff(second,in_time,out_time))/count(distinct uid),1) avg_viiew_len_sec
from tb_user_log
where artical_id != 0
and date_format(in_time,'%Y-%m') = '2021-11'
group by 1
order by 2;