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Asked:2022-04-14 00:13:27 +0000 UTC2022-04-14 00:13:27 +0000 UTC

pafy.get_playlist 抛出错误

  • 772

为什么会出现错误？

import pafy


def get_playlist(url):
    playlist = pafy.get_playlist(url)
    return playlist['items']


if __name__ == '__main__':
    yt_url = input('Введите ссылку: ')
    p = get_playlist(yt_url)

关联： https://www.youtube.com/playlist?list=PL-Sp30dZBoZ2nBqUkUB3C93YHiVJCp-GY

错误：

Traceback (most recent call last):
  File "D:/pyprojects/yt_dl/test2.py", line 11, in <module>
    p = get_playlist(yt_url)
  File "D:/pyprojects/yt_dl/test2.py", line 5, in get_playlist
    playlist = pafy.get_playlist(url)
  File "D:\pyprojects\venv\lib\site-packages\pafy\playlist.py", line 58, in get_playlist
    allinfo = fetch_decode(url)  # unicode
  File "D:\pyprojects\venv\lib\site-packages\pafy\pafy.py", line 67, in fetch_decode
    req = g.opener.open(url)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 531, in open
    response = meth(req, response)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 640, in http_response
    response = self.parent.error(
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 563, in error
    result = self._call_chain(*args)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 502, in _call_chain
    result = func(*args)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 755, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 531, in open
    response = meth(req, response)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 640, in http_response
    response = self.parent.error(
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 569, in error
    return self._call_chain(*args)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 502, in _call_chain
    result = func(*args)
  File "C:\Users\Admin\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 649, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found
python

1 个回答

  1. Best Answer

    据我了解，该库pafy非常旧，自 2014 年以来一直没有更新，并且不适用于当前版本的Youtube. 然而，在内部，这个库使用了一个完全最新的库youtube_dl，你可以使用它：

    import youtube_dl

ydl_opts = {}
with youtube_dl.YoutubeDL(ydl_opts) as ydl:
    playlist = ydl.extract_info("https://www.youtube.com/playlist?list=PL634F2B56B8C346A2", download=False)
    print(playlist['title'])

    结论（我把下载信息的服务信息缩短了，留下播放列表标题的实际输出）：

    Rick Astley playlist

    如果download=False在调用该函数时去掉该参数extract_info，则播放列表中列出的所有视频都将下载到当前目录。

    • 2

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