KeyError：＆＃x27; list＆＃x27;使用Pytube向用户询问输入（）的URL之后

Question

我尝试在YouTube中下载整个播放列表并使用此代码。

我还更新了Pytube，pip，python，并尝试更改代码，但我不知道如何。

这是我第一次编码，我需要一个好的YouTube转换器。

from pytube.contrib.playlist import Playlist
from pytube import YouTube
from pytube.cli import on_progress

url = input("https://youtube.com/playlist?list=PLVxmNUcoG69uwwG59UVUyJ9SoLRHE78kp")
playlist = Playlist(url)
print("Total Videos: ", len(playlist.video_urls))

for video_url in playlist.video_urls:
    yt = YouTube(video_url, on_progress_callback=on_progress)
    stream = yt.streams.get_highest_resolution()
    print(yt.title)
    stream.download(filename=yt.title + ".mp4")

这是我的错误。当我在链接显示后链接插入文本后，单击Enter时。

https://youtube.com/playlist?list=PLVxmNUcoG69uwwG59UVUyJ9SoLRHE78kp
-- Click Enter

Traceback (most recent call last):
  File "/Users/jonasmaier/Desktop/downi you/codes/playlist.py", line 7, in <module>
    print("Total Videos: ",len(playlist.video_urls))
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/helpers.py", line 89, in __len__
    self.generate_all()
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/helpers.py", line 105, in generate_all
    next_item = next(self.gen)
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/contrib/playlist.py", line 281, in url_generator
    for page in self._paginate():
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/contrib/playlist.py", line 118, in _paginate
    json.dumps(extract.initial_data(self.html))
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/contrib/playlist.py", line 58, in html
    self._html = request.get(self.playlist_url)
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/contrib/playlist.py", line 48, in playlist_url
    return f"https://www.youtube.com/playlist?list={self.playlist_id}"
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/contrib/playlist.py", line 39, in playlist_id
    self._playlist_id = extract.playlist_id(self._input_url)
  File "/Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/site-packages/pytube/extract.py", line 151, in playlist_id
    return parse_qs(parsed.query)['list'][0]
KeyError: 'list'

Answer 1

  url = input（“ https://youtube.com/playlist?list=plvxmnucog69uwwg59uvuyj9solrhe78kp”）
 

如果仅按Enter，这并不意味着url现在是字符串“ https://youtube.com/playlist?list=plvxmnucog69uwwg59uwwg59uvuyj9solrhe78kp”。

相反，input（）函数将URL显示为a 提示向用户显示并返回用户输入的内容，因此在这种情况下：没有。因此，url实际上是一个空字符串。因此，Pytube无法从（空）URL解析播放列表ID。

您的意思是：

url = input("Enter the URL: ")

然后当程序运行时，您必须输入URL并按Enter。

或者：

url = "https://youtube.com/playlist?list=PLVxmNUcoG69uwwG59UVUyJ9SoLRHE78kp"

程序只使用固定的URL而不询问用户。