how to get name of a file in directory using python

Question

There is an mkv file in a folder named "export". What I want to do is to make a python script which fetches the file name from that export folder. Let's say the folder is at "C:\Users\UserName\Desktop\New_folder\export".

How do I fetch the name?

I tried using this os.path.basename and os.path.splitext .. well.. didn't work out like I expected.

Answer 1

os.path implements some useful functions on pathnames. But it doesn't have access to the contents of the path. For that purpose, you can use os.listdir.

The following command will give you a list of the contents of the given path:

os.listdir("C:\Users\UserName\Desktop\New_folder\export")

Now, if you just want .mkv files you can use fnmatch(This module provides support for Unix shell-style wildcards) module to get your expected file names:

import fnmatch
import os

print([f for f in os.listdir("C:\Users\UserName\Desktop\New_folder\export") if fnmatch.fnmatch(f, '*.mkv')])

Also as @Padraic Cunningham mentioned as a more pythonic way for dealing with file names you can use glob module :

map(path.basename,glob.iglob(pth+"*.mkv"))

Answer 2

You can use glob:

from glob import glob

pth ="C:/Users/UserName/Desktop/New_folder/export/"
print(glob(pth+"*.mkv"))

path+"*.mkv" will match all the files ending with .mkv.

To just get the basenames you can use map or a list comp with iglob:

from glob import iglob

print(list(map(path.basename,iglob(pth+"*.mkv"))))


print([path.basename(f) for f in  iglob(pth+"*.mkv")])

iglob returns an iterator so you don't build a list for no reason.