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 document icon in Explorer or Finder. What is the best way to do this in Python?

Answer 1

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.

Answer 2

open and start are command-interpreter things for Mac OS/X and Windows respectively, to do this.

To call them from Python, you can either use subprocess module or os.system().

Here are considerations on which package to use:

1. You can call them via os.system, which works, but...

   Escaping: os.system only works with filenames that don't have any spaces or other shell metacharacters in the pathname (e.g. A:\abc\def\a.txt), or else these need to be escaped. There is shlex.quote for Unix-like systems, but nothing really standard for Windows. Maybe see also python, windows : parsing command lines with shlex

   • MacOS/X: os.system("open " + shlex.quote(filename))
   • Windows: os.system("start " + filename) where properly speaking filename should be escaped, too.

2. You can also call them via subprocess module, but...

   For Python 2.7 and newer, simply use

   subprocess.check_call(['open', filename])
   

   In Python 3.5+ you can equivalently use the slightly more complex but also somewhat more versatile

   subprocess.run(['open', filename], check=True)
   

   If you need to be compatible all the way back to Python 2.4, you can use subprocess.call() and implement your own error checking:

   try:
       retcode = subprocess.call("open " + filename, shell=True)
       if retcode < 0:
           print >>sys.stderr, "Child was terminated by signal", -retcode
       else:
           print >>sys.stderr, "Child returned", retcode
   except OSError, e:
       print >>sys.stderr, "Execution failed:", e
   

   Now, what are the advantages of using subprocess?

   • Security: In theory, this is more secure, but in fact we're needing to execute a command line one way or the other; in either environment, we need the environment and services to interpret, get paths, and so forth. In neither case are we executing arbitrary text, so it doesn't have an inherent "but you can type 'filename ; rm -rf /'" problem, and if the file name can be corrupted, using subprocess.call gives us little additional protection.
   • Error handling: It doesn't actually give us any more error detection, we're still depending on the retcode in either case; but the behavior to explicitly raise an exception in the case of an error will certainly help you notice if there is a failure (though in some scenarios, a traceback might not at all be more helpful than simply ignoring the error).
   • Spawns a (non-blocking) subprocess: We don't need to wait for the child process, since we're by problem statement starting a separate process.

   To the objection "But subprocess is preferred." However, os.system() is not deprecated, and it's in some sense the simplest tool for this particular job. Conclusion: using os.system() is therefore also a correct answer.

   A marked disadvantage is that the Windows start command requires you to pass in shell=True which negates most of the benefits of using subprocess.

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

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

Answer 5

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])

Answer 6

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 7

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.

Answer 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.

Answer 9

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 10

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))

Answer 11

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.

Answer 12

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))

Answer 13

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]")

Answer 14

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.

Answer 15

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

$ pip install universal-startfile

then launch a file or URL:

from startfile import startfile

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

Answer 16

I think you might want to open file in editor.

For Windows

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

For Linux

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

Answer 17

I was getting an error when calling my open file() function. I was following along with a guide but the guide was written in windows while I'm on Linux. So the os.statrfile method wasn't working for me. I was able to alleviate this problem by doing the following:

Import libraries

import sys, os, subprocess
import tkinter
import tkinter.filedioalog as fd
import tkinter.messagebox as mb

After the lib imports I then called the subprocess method for opening a file in unix based OS which is "xdg-open" and the file that will be opened.

def open_file():
     file = fd.askopenfilename(title='Choose a file of any type', filetypes=[('All files', "*.*")])
     subprocess.call(['xdg-open', file])