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') as dt,
       round(sum(timestampdiff(second,in_time,out_time))/count(distinct uid),1) as avg_viiew_len_sec
from tb_user_log as a
where date_format(in_time,'%Y-%m')='2021-11' and artical_id != 0
group by date_format(in_time,'%Y-%m-%d')
order by avg_viiew_len_sec
每日的总浏览时长/每天的观看人数；这里要用到distinct去重 用户可能一个文章一天内看了多次
sum(timestampdiff(second,in_time,out_time))/count(distinct uid)