题解 | #2021年11月每天的人均浏览文章时长#

select day,
        round(sum(second_sum)/count(uid),1) time_avg
        from
(select substr(in_time,1,10) day,
        uid,
        sum(timestampdiff(second,in_time,out_time)) second_sum
        from tb_user_log
        where substr(in_time,1,10) >='2021-11-01'
        and artical_id !=0
        group by substr(in_time,1,10),uid
) a
group by day
order by round(sum(second_sum)/count(uid),1)