This script is for exporting your playlist items to a single folder. Save, change two variables to suit your needs and run =) One use case is: you have a music collection and a playlist containing the best songs of this collection, and you want to load only the best songs to your MP3 player with limited space. This also doesn't overwrite files, …

export.py

import os
import shutil

directory_name = "playlist/"
playlist_name = "playlist.m3u"

f = open(playlist_name, "r")
files = []
for line in f:
    line = line.strip().strip("\n").strip("\r")
    if line and not line.startswith("#"):
        files.append(line)
dir_contents = os.listdir(directory_name)
for file in files:
    try:
        filename = os.path.basename(file)
        if filename not in dir_contents:
            shutil.copy(file, directory_name)
            print filename
        else:
            print ".", #File already there
    except Exception as e:
        print str(e)

Nice script :)

Suggestion: Even better if you can assign a numerical prefix to the mp3 file names (e.g 001, 002,...) to preserve the oder of the song in the playlist.

-bodhi

Hey @mibodhi ! This wasn't needed for me, but you can certainly do this if you change line 14 to for i, file in enumerate(files): and line 16 to filename = str(i).zfill(3)+"_"+os.path.basename(file). Hope this helps! -Arsenijs

Cool! I'll give it a try.
-bodhi

I think line 18 still uses the original name. So we also need to somehow to rename the current file to filename:
shutil.copy(file, directory_name)

-bodhi

Indeed! Add file = os.path.join(os.path.dirname(file), filename) after line 16.