2021年11月每天的人均浏览文章时长

id	uid	artical_id	in_time	out_time	sign_cin
1	101	9001	2021-11-01 10:00:00	2021-11-01 10:00:31	0
2	102	9001	2021-11-01 10:00:00	2021-11-01 10:00:24	0
3	102	9002	2021-11-01 11:00:00	2021-11-01 11:00:11	0
4	101	9001	2021-11-02 10:00:00	2021-11-02 10:00:50	0
5	102	9002	2021-11-02 11:00:01	2021-11-02 11:00:24	0

（uid-用户ID, artical_id-文章ID, in_time-进入时间, out_time-离开时间, sign_in-是否签到）

场景逻辑说明：artical_id-文章ID代表用户浏览的文章的ID，artical_id-文章ID为0表示用户在非文章内容页（比如App内的列表页、活动页等）。

问题：统计2021年11月每天的人均浏览文章时长（秒数），结果保留1位小数，并按时长由短到长排序。

输出示例：

示例数据的输出结果如下

dt	avg_viiew_len_sec
2021-11-01	33.0
2021-11-02	36.5

解释：

11月1日有2个人浏览文章，总共浏览时长为31+24+11=66秒，人均浏览33秒；

11月2日有2个人浏览文章，总共时长为50+23=73秒，人均时长为36.5秒。

示例1

输入:

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid INT NOT NULL COMMENT '用户ID',
    artical_id INT NOT NULL COMMENT '视频ID',
    in_time datetime COMMENT '进入时间',
    out_time datetime COMMENT '离开时间',
    sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;

INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
  (101, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:31', 0),
  (102, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:24', 0),
  (102, 9002, '2021-11-01 11:00:00', '2021-11-01 11:00:11', 0),
  (101, 9001, '2021-11-02 10:00:00', '2021-11-02 10:00:50', 0),
  (102, 9002, '2021-11-02 11:00:01', '2021-11-02 11:00:24', 0);

输出:

2021-11-01|33.0
2021-11-02|36.5

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

Mysql 解法, 执行用时: 36ms, 内存消耗: 6368KB, 提交时间: 2022-01-25

select date_format(in_time,"%Y-%m-%d")dt,
round (sum(timestampdiff(second,in_time,out_time))/count(distinct uid),1)avg_view_len_sec
from tb_user_log
where month(in_time)= 11 and artical_id !=0
group by dt
order by avg_view_len_sec

Mysql 解法, 执行用时: 36ms, 内存消耗: 6404KB, 提交时间: 2021-12-08

select DATE_FORMAT(in_time,'%Y-%m-%d') as a,
round(sum(TIMESTAMPDIFF(second,in_time,out_time))/count(distinct uid),1) as b
from tb_user_log
where year(in_time)=2021 and month(in_time)=11 and artical_id!=0
group by a
order by b ;

Mysql 解法, 执行用时: 36ms, 内存消耗: 6416KB, 提交时间: 2022-01-25

select date_format(in_time,'%Y-%m-%d') dt,
       round(sum(timestampdiff(second,in_time,out_time)) /count(distinct uid),1) avg_view_len_sec
from tb_user_log
where date_format(in_time,'%Y-%m')='2021-11' and artical_id != 0
group by dt
order by avg_view_len_sec

Mysql 解法, 执行用时: 36ms, 内存消耗: 6416KB, 提交时间: 2021-12-08

select dt,round(sum(sc)/count(distinct uid),1) avg_view_len_sec
from 
(select uid,date(out_time) dt,TIMESTAMPDIFF(SECOND,in_time,out_time) sc,artical_id from tb_user_log) t 
where year(dt) = "2021" and month(dt) = "11" and artical_id != 0
group by dt
order by avg_view_len_sec ;

Mysql 解法, 执行用时: 36ms, 内存消耗: 6424KB, 提交时间: 2022-02-10

-- 人均浏览文章时长 timestampdiff(second,out_time,in_time)
select left(in_time,10)as dt,
round(sum(timestampdiff(second,in_time,out_time))/count(distinct uid),1) as avg_view_len_sec
from tb_user_log
where artical_id<>0
and in_time like '2021-11%'
group BY dt
order by avg_view_len_sec

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