Get the Firefox Default Profile

To get the Firefox default profile, I am using the following program.

import glob
from pathlib import Path
import re

x = glob.glob(str(Path.home()) + '/.mozilla/firefox/*/')
r = re.compile(".*default-release.*")
firefox_default_profile = list(filter(r.match, x))[0]
print(firefox_default_profile)

The logic is simple. The path looks like ~/.mozilla/firefox/5ierwkox.default-release.

In the whole path the only variable is the 5ierwkox. It varies from person to person.

The program works fine, but I am wondering if I have over-complicated the code or not.

• You don’t need to import glob; pathlib.Path has a Path.glob method.
• You don’t need to import re; you can merge the regex into the glob – .*default-release.* become *default-release*.
• You can use next(...) rather than list(...)[0].
• By passing None as a second argument to next we can prevent the code from erroring if the user doesn’t have a Firefox profile.

import pathlib

profiles = pathlib.Path.home().glob(".mozilla/firefox/*default-release*/")
firefox_default_profile = next(profiles, None)
print(firefox_default_profile)

None of my profiles have default-release in the name.

To find the default profile, we need to parse ~/.mozilla/firefox/profiles.ini and find the section with Default=1 set.