How to open a file with the standard application?

Question

My application prints a PDF to a temporary file. How can I open that file with the default application in Python?

I need a solution for

• Windows
• Linux (Ubuntu with Xfce if there's nothing more general.)

Related

• Open document with default application in Python

Answer 1

if linux:
    os.system('xdg-open "$file"') #works for urls too
else:
    os.system('start "$file"') #a total guess

Answer 2

on windows it works with os.system('start <myFile>'). On Mac (I know you didn't ask...) it's os.system('open <myFile>')

Answer 3

os.startfile is only available for windows for now, but xdg-open will be available on any unix client running X.

if sys.platform == 'linux2':
    subprocess.call(["xdg-open", file])
else:
    os.startfile(file)

Answer 4

Ask your favorite Application Framework for how to do this in Linux.

This will work on Windos and Linux as long as you use GTK:

import gtk
gtk.show_uri(gtk.gdk.screen_get_default(), URI, 0)

where URI is the local URL to the file

Answer 5

Open file using an application that your browser thinks is an appropriate one:

import webbrowser
webbrowser.open_new_tab(filename)

Answer 6

A small correction is necessary for NicDumZ's solution to work exactly as given. The problem is with the use of 'is' operator. A working solution is:

if sys.platform == 'linux2':
    subprocess.call(["xdg-open", file])
else:
    os.startfile(file)

A good discussion of this topic is at http://stackoverflow.com/questions/132988/is-there-a-difference-between-and-is-in-python.