SQL面试题（二）

原题参考自：SQL 大厂面试真题篇_sql大厂面试题_withoutfear的博客-优快云博客

SQL1 各个视频的平均完播率

描述: 用户-视频互动表tb_user_video_log

id	uid	video_id	start_time	end_time	if_follow	if_like	if_retweet	comment_id
1	101	2001	2021-10-01  10:00:00	2021-10-01  10:00:30	0	1	1	NULL
2	102	2001	2021-10-01  10:00:00	2021-10-01  10:00:24	0	0	1	NULL
3	103	2001	2021-10-01  11:00:00	2021-10-01  11:00:34	0	1	0	1732526
4	101	2002	2021-09-01  10:00:00	2021-9-01  10:00:42	1	0	1	NULL
5	102	2002	2021-10-01  11:00:00	2021-10-01  10:00:30	1	0	1	NULL

（uid-用户ID, video_id-视频ID, start_time-开始观看时间, end_time-结束观看时间, if_follow-是否关注, if_like-是否点赞, if_retweet-是否转发, comment_id-评论ID）

短视频信息表tb_video_info

id	video_id	author	tag	duration	release_time
1	2001	901	影视	30	2021-01-01 07:00:00
2	2002	901	美食	60	2021-01-01 07:00:00
3	2003	902	旅游	90	2021-01-01 07:00:00

（video_id-视频ID, author-创作者ID, tag-类别标签, duration-视频时长（秒）, release_time-发布时间）

问题：计算2021年里有播放记录的每个视频的完播率(结果保留三位小数)，并按完播率降序排序

注：视频完播率是指完成播放次数占总播放次数的比例。简单起见，结束观看时间与开始播放时间的差>=视频时长时，视为完成播放。
本人写的：

select 
	video_id,
	round(sum(com_stat)/count(video_id),3) as com_ro
from
	(
	select
		aa.video_id,
		start_time,
		end_time,
		timestampdiff(second,end_time,start_time) as play_time,
		case when play_time>=bb.duration then 1 else 0 end as com_stat
	from
		tb_user_video_log aa
	left join 
		tb_video_info bb
	on
		aa.video_id=bb.video_id
	where 
		year(start_time)=2021
	)
group by video_id
order by com_ro desc;

chatGPT 3.5：

SELECT 
	l.video_id, 
	ROUND(SUM(CASE WHEN TIMESTAMPDIFF(SECOND, l.start_time, l.end_time) >= v.duration THEN 1 ELSE 0 END) / COUNT(*), 3) AS completion_rate
FROM 
	tb_user_video_log l
JOIN 
	tb_video_info v 
ON 
	l.video_id = v.video_id
WHERE 
	YEAR(l.start_time) = 2021
GROUP BY l.video_id
ORDER BY completion_rate DESC;

问题：计算各类视频的平均播放进度，将进度大于60%的类别输出。

注：播放进度=播放时长÷视频时长*100%，当播放时长大于视频时长时，播放进度均记为100%。

结果保留两位小数，并按播放进度倒序排序。

本人写的：

select 
	tag,
	round(avg(play_per),2) as avg_play_per
from
	(	
	select
		aa.video_id,
		aa.start_time,
		aa.end_time,
		timestampdiff(second,aa.end_time,aa.start_time) as play_time,
		if(play_time>bb.duration,1,play_time/bb.duration)*100 as play_per
		bb.tag
	from
		tb_user_video_log aa
	left join 
		tb_video_info bb
	on
		aa.video_id=bb.video_id
	)
group by tag
having avg_play_per>60
order by avg_play_per desc;

chatGPT：

SELECT 
	v.tag,
	ROUND(SUM(CASE WHEN TIMESTAMPDIFF(SECOND, l.start_time, l.end_time) >= v.duration THEN 100 ELSE TIMESTAMPDIFF(SECOND, l.start_time, l.end_time) / v.duration * 100 END) / COUNT(*), 2) AS avg_progress
FROM 
	tb_user_video_log l
JOIN 
	tb_video_info v 
ON 
	l.video_id = v.video_id
GROUP BY v.tag
HAVING avg_progress > 60
ORDER BY avg_progress DESC;