给出诸如"文件:///Volumes/Shared/1-05%20Born%20to%20be%20Wild.m4a"之类的URI。什么是最简单的方法将其转换为" / Volumes / Shared / 1-05天生就是Wild.m4a"所以我可以将它作为参数传递给open()或类似的东西?
答案 0 :(得分:0)
看起来这应该做你想要的:
import urllib2
url = "file:///Volumes/Shared/1-05%20Born%20to%20be%20Wild.m4a"
if url[0:7] == 'file://':
cleaned_url = url[7:]
cleaned_url = urllib2.unquote(url_string)