我正在尝试使用 youtube-dl 从 youtube 的播放列表列表中获取一些信息。我已经编写了这段代码,但它需要的不是视频信息而是播放列表信息(例如播放列表标题而不是播放列表中的视频标题)。我不明白为什么。
input_file = open("url")
for video in input_file:
print(video)
ydl_opts = {
'ignoreerrors': True
}
with youtube_dl.YoutubeDL(ydl_opts) as ydl:
info_dict = ydl.extract_info(video, download=False)
for i in info_dict:
video_thumbnail = info_dict.get("thumbnail"),
video_id = info_dict.get("id"),
video_title = info_dict.get("title"),
video_description = info_dict.get("description"),
video_duration = info_dict.get("duration")
任何帮助将不胜感激。
您调用的变量
video
实际上保存的是播放列表信息,而不是视频信息。您可以在播放列表的 entries
属性中找到各个视频信息的列表。
请参阅下文了解可能的修复方法。我将您的
video
变量重命名为 playlist
并自由地重写它并添加输出:
import textwrap
import youtube_dl
playlists = [
"https://www.youtube.com/playlist?list=PLRQGRBgN_EnrPrgmMGvrouKn7VlGGCx8m"
]
for playlist in playlists:
with youtube_dl.YoutubeDL({"ignoreerrors": True, "quiet": True}) as ydl:
playlist_dict = ydl.extract_info(playlist, download=False)
# Pretty-printing the video information (optional)
for video in playlist_dict["entries"]:
print("\n" + "*" * 60 + "\n")
if not video:
print("ERROR: Unable to get info. Continuing...")
continue
for prop in ["thumbnail", "id", "title", "description", "duration"]:
print(prop + "\n" +
textwrap.indent(str(video.get(prop)), " | ", lambda _: True)
)
运行命令
youtube-dl --print-json https://www.youtube.com/playlist?list=<playlist_id> > example.json
您还可以使用
--get
来检索特定项目,例如
youtube-dl --get-title https://www.youtube.com/playlist?list=<playlist_id> > example.txt