Convert Local File URL to File Path-CodePudding

I have a URL that points to a local file.

'file:///home/pi/Desktop/music/Radio Song.mp3'

I need to somehow convert this into a traditional file path, like the os module employs.

'/home/pi/Desktop/music/Radio Song.mp3'

Right now I'm hacking it with the replace() method.

path = file.replace('file://', '').replace(' ', ' ')

I've looked at the os module, and it doesn't seem to have support for this. I've searched various ways of phrasing it, and I can't seem to find an answer. Am I just ignorant of the terminology? What's the proper way to do this?

CodePudding user response：

The following would work:

from urllib.request import url2pathname
from urllib.parse import urlparse

p = urlparse('file:///home/pi/Desktop/music/Radio Song.mp3')
file_path = url2pathname(p.path)
print(file_path)

(thanks to user @MillerTime correctly pointing out that the solution will not remove file:// without the urlparse)

Output:

/home/pi/Desktop/music/Radio Song.mp3

urllib is a standard library, so no installation required.

On a Windows machine, running just url2pathname would give you a valid file path right away, but it would be relative to the working directory of the script - e.g. running it from somewhere on D: drive:

D:\home\pi\Desktop\music\Radio Song.mp3