题解 | #2021年11月每天的人均浏览文章时长#

SELECT date(in_time) dt, round(sum(time_to_sec(timediff(out_time,in_time)))/count(DISTINCT uid),1) AS avg_viiew_len_sec FROM tb_user_log
WHERE substr(date(in_time),1,7) = '2021-11' AND artical_id <> 0
GROUP BY dt
ORDER BY avg_viiew_len_sec;