题解 | #2021年11月每天的人均浏览文章时长#

select
    substr(in_time,1,10) as dt,
    round(sum(TIMESTAMPDIFF(SECOND, in_time, out_time)) / count(distinct uid), 1) as avg_viiew_len_sec
from tb_user_log
where substr(in_time,1,7) = '2021-11'
and artical_id != 0
group by 
    substr(in_time,1,10)
order by round(sum(TIMESTAMPDIFF(SECOND, in_time, out_time)) / count(distinct uid), 1) asc