How do I know what program 'xdg-open' will use to open a directory? (Python)

Asked Aug 25 '15 at 17:14

Active Aug 25 '15 at 17:14

Viewed 798 times

So I have this:

def openSegmentDirectory(self):
    if platform.system() == "Windows":
        os.startfile(self.segmentPath)
    elif platform.system() == "Linux":
        subprocess.Popen(['xdg-open', self.segmentPath])
    else:
        #for mac
        os.system('open "%s"' % self.segmentPath)

But when using it on linux, the system attempts to open the directory in firefox, which I don't want. How can I change this?

asked Aug 25 '15 at 17:14

pretzlstyle

2,774
5
23
40

You might want to look at xdg-mime default option -- http://portland.freedesktop.org/xdg-utils-1.0/xdg-mime.html – clockwatcher Aug 25 '15 at 17:29
But is this changed on a local system level? I need something to put into my python code to ensure it works on any system that may run it. – pretzlstyle Aug 25 '15 at 17:33
It uses your default program for opening files – Padraic Cunningham Aug 25 '15 at 17:35
But I have Dolphin and several other file managers installed, and I never get Firefox unless I do this in python with this code. It is opening Firefox as a last resort and I don't know why. I'm using RHEL – pretzlstyle Aug 25 '15 at 17:38
So `xdg-open /home` from a shell does not use firefox? – Padraic Cunningham Aug 25 '15 at 17:53
No, you''re right, it also acts this way from terminal. Even so, I need a universal solution, since any machine may have this issue. – pretzlstyle Aug 25 '15 at 18:26

How do I know what program 'xdg-open' will use to open a directory? (Python)

0 Answers0