file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3
如何转换像上面这样的字符串来获取我可以传递给open()
函数的普通文件路径?
看看url2pathname
:
import urllib2
path = urllib2.url2pathname("file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3")
这被称为unquote。可从urllib获得。
import urllib
urllib.unquote('%20')