题解 | 2021年11月每天的人均浏览文章时长

select 
    dt,
    round(sum(view_len)/count(distinct uid),1) avg_viiew_len_sec
from(
    select
        date(in_time) dt,
        uid,
        if(artical_id=0,0,timestampdiff(second,in_time,out_time)) view_len
    from
        tb_user_log
    where
        date(in_time) between '2021-11-01' and '2021-11-30'
    ) a
group by
    1
order by 
    avg_viiew_len_sec