题解 | #2021年11月每天的人均浏览文章时长#

select time, avg_viiew_len_sec
from
(select time, round(sum(timestampdiff(second, in_time, out_time))/count(distinct uid),1) as avg_viiew_len_sec
from
(select uid, artical_id, in_time, out_time, date_format(in_time, '%Y-%m-%d') as time
from tb_user_log
where artical_id !=0) as t1
where time >='2021-11-01' and time<='2021-11-03'
group by time) as t2
order by avg_viiew_len_sec