Open document with default OS application in Python, both in Windows and Mac OS

Question

I need to be able to open a document using its default application in Windows and Mac OS. Basically, I want to do the same thing that happens when you double-click on the do

Answer 1

Start does not support long path names and white spaces. You have to convert it to 8.3 compatible paths.

import subprocess
import win32api

filename = "C:\\Documents and Settings\\user\\Desktop\file.avi"
filename_short = win32api.GetShortPathName(filename)

subprocess.Popen('start ' + filename_short, shell=True )

The file has to exist in order to work with the API call.

Answer 2

Just for completeness (it wasn't in the question), xdg-open will do the same on Linux.

Answer 3

I prefer:

os.startfile(path, 'open')

Note that this module supports filenames that have spaces in their folders and files e.g.

A:\abc\folder with spaces\file with-spaces.txt

(python docs) 'open' does not have to be added (it is the default). The docs specifically mention that this is like double-clicking on a file's icon in Windows Explorer.

This solution is windows only.

Answer 4

on mac os you can call 'open'

import os
os.popen("open myfile.txt")

this would open the file with TextEdit, or whatever app is set as default for this filetype

Answer 5

If you have to use an heuristic method, you may consider webbrowser.
It's standard library and despite of its name it would also try to open files:

Note that on some platforms, trying to open a filename using this function, may work and start the operating system’s associated program. However, this is neither supported nor portable. (Reference)

I tried this code and it worked fine in Windows 7 and Ubuntu Natty:

import webbrowser
webbrowser.open("path_to_file")

This code also works fine in Windows XP Professional, using Internet Explorer 8.

Answer 6

If you want to specify the app to open the file with on Mac OS X, use this: os.system("open -a [app name] [file name]")