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

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?

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.

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.