题解 | #SQL162 2021年11月每天的人均浏览文章时长#
2021年11月每天的人均浏览文章时长
https://www.nowcoder.com/practice/8e33da493a704d3da15432e4a0b61bb3
WITH t1 AS ( SELECT DATE(in_time) AS dt, SUM(TIMESTAMPDIFF(SECOND, in_time, out_time)) AS sum_viiew_len_sec, COUNT(DISTINCT uid) AS total FROM tb_user_log WHERE DATE (in_time) BETWEEN '2021-11-01' AND '2021-11-30' AND artical_id != 0 GROUP BY dt ) SELECT dt, ROUND(sum_viiew_len_sec / total,1) AS avg_viiew_len_sec FROM t1 ORDER BY avg_viiew_len_sec ASC; # 关键在于时间筛选和文章id不为0