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

On mac os you can call open:

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

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

I prefer:

os.startfile(path, 'open')

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

A:abcfolder with spacesfile 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.

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

Use the subprocess module available on Python 2.4+, not os.system(), so you don’t have to deal with shell escaping.

import subprocess, os, platform
if platform.system() == 'Darwin':       # macOS
    subprocess.call(('open', filepath))
elif platform.system() == 'Windows':    # Windows
    os.startfile(filepath)
else:                                   # linux variants
    subprocess.call(('xdg-open', filepath))

The double parentheses are because subprocess.call() wants a sequence as its first argument, so we’re using a tuple here. On Linux systems with Gnome there is also a gnome-open command that does the same thing, but xdg-open is the Free Desktop Foundation standard and works across Linux desktop environments.

import os
import subprocess

def click_on_file(filename):
    '''Open document with default application in Python.'''
    try:
        os.startfile(filename)
    except AttributeError:
        subprocess.call(['open', filename])

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]")

If you want to go the subprocess.call() way, it should look like this on Windows:

import subprocess
subprocess.call(('cmd', '/C', 'start', '', FILE_NAME))

You can’t just use:

subprocess.call(('start', FILE_NAME))

because start is not an executable but a command of the cmd.exe program. This works:

subprocess.call(('cmd', '/C', 'start', FILE_NAME))

but only if there are no spaces in the FILE_NAME.

While subprocess.call method enquotes the parameters properly, the start command has a rather strange syntax, where:

start notes.txt

does something else than:

start "notes.txt"

The first quoted string should set the title of the window. To make it work with spaces, we have to do:

start "" "my notes.txt"

which is what the code on top does.

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\Desktopfile.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.

I am pretty late to the lot, but here is a solution using the windows api. This always opens the associated application.

import ctypes

shell32 = ctypes.windll.shell32
file = 'somedocument.doc'

shell32.ShellExecuteA(0,"open",file,0,0,5)

A lot of magic constants. The first zero is the hwnd of the current program. Can be zero. The other two zeros are optional parameters (parameters and directory). 5 == SW_SHOW, it specifies how to execute the app.
Read the
ShellExecute API docs for more info.

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.

os.startfile(path, 'open') under Windows is good because when spaces exist in the directory, os.system('start', path_name) can’t open the app correctly and when the i18n exist in the directory, os.system needs to change the unicode to the codec of the console in Windows.

On windows 8.1, below have worked while other given ways with subprocess.call fails with path has spaces in it.

subprocess.call('cmd /c start "" "any file path with spaces"')

By utilizing this and other’s answers before, here’s an inline code which works on multiple platforms.

import sys, os, subprocess
subprocess.call(('cmd /c start "" "'+ filepath +'"') if os.name is 'nt' else ('open' if sys.platform.startswith('darwin') else 'xdg-open', filepath))

Here is the answer from Nick, adjusted slightly for WSL:

import os
import sys
import logging
import subprocess

def get_platform():
    if sys.platform == 'linux':
        try:
            proc_version = open('/proc/version').read()
            if 'Microsoft' in proc_version:
                return 'wsl'
        except:
            pass
    return sys.platform

def open_with_default_app(filename):
    platform = get_platform()
    if platform == 'darwin':
        subprocess.call(('open', filename))
    elif platform in ['win64', 'win32']:
        os.startfile(filename.replace('/','\'))
    elif platform == 'wsl':
        subprocess.call('cmd.exe /C start'.split() + [filename])
    else:                                   # linux variants
        subprocess.call(('xdg-open', filename))

I think you might want to open file in editor.

For Windows

subprocess.Popen(["notepad", filename])

For Linux

subprocess.Popen(["text-editor", filename])

I built a small library combining the best options for cross-platform support:

$ pip install universal-startfile

then

from startfile import startfile

startfile("~/Downloads/example.png")
startfile("http://example.com")