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Python 我尝试使用Paffy获取YouTube播放列表的视频链接时出错_Python_Python 3.x_Pafy - Fatal编程技术网

Python 我尝试使用Paffy获取YouTube播放列表的视频链接时出错

Python 我尝试使用Paffy获取YouTube播放列表的视频链接时出错,python,python-3.x,pafy,Python,Python 3.x,Pafy,在获得正确的输出后,它向我显示一个错误,即列表索引超出范围 有人能帮我避免这个错误吗 “这里只是猜测,但您可能应该将l=len(播放列表['author'])更改为l=len(播放列表['items']),因为您希望循环播放列表项 import pafy plurl = input("Enter The Link : ") playlist = pafy.get_playlist(plurl) l = len(playlist['author']) #file = open("VideoInf

在获得正确的输出后,它向我显示一个错误,即列表索引超出范围 有人能帮我避免这个错误吗


这里只是猜测,但您可能应该将
l=len(播放列表['author'])
更改为
l=len(播放列表['items'])
,因为您希望循环播放列表项

import pafy
plurl = input("Enter The Link : ")
playlist = pafy.get_playlist(plurl)
l = len(playlist['author'])

#file = open("VideoInformation.txt","w+")

def extract_name_from_title(s):
c=0
name=""
start = s.find("-")
start = start+2
start = int(start)
end = s.find("|")
end = end-1
end = int(end)
name = s[start:end]

return name

for i in range(0,l-1) :
  id = playlist['items'][i]['pafy'].videoid
  v=pafy.new(id)
  #video = pafy.new('https://www.youtube.com/watch?v='+id)
  print("Video Number : {}".format(i))
  name = extract_name_from_title(v.title)
  link = "https://www.youtube.com/watch?v={}".format(id)
  print("Video title :{}".format(name))
  print("Video Link : {}".format(link))