Question

I want to open a process in the background and interact with it, but this process should be invisible in both Linux and Windows. In Windows you have to do some stuff with STARTUPINFO, while this isn't valid in Linux:

ValueError: startupinfo is only supported on Windows platforms

Is there a simpler way than creating a separate Popen command for each OS?

if os.name == 'nt':
    startupinfo = subprocess.STARTUPINFO()
    startupinfo.dwFlags |= subprocess.STARTF_USESHOWWINDOW
    proc = subprocess.Popen(command, startupinfo=startupinfo)
if os.name == 'posix':
    proc = subprocess.Popen(command)    
Was it helpful?

Solution

You can reduce one line :)

startupinfo = None
if os.name == 'nt':
    startupinfo = subprocess.STARTUPINFO()
    startupinfo.dwFlags |= subprocess.STARTF_USESHOWWINDOW
proc = subprocess.Popen(command, startupinfo=startupinfo)

OTHER TIPS

Just a note: for Python 2.7 I have to use subprocess._subprocess.STARTF_USESHOWWINDOW instead of subprocess.STARTF_USESHOWWINDOW.

I'm not sure you can get much simpler than what you've done. You're talking about optimising out maybe 5 lines of code. For the money I would just get on with my project and accept this as a consquence of cross-platform development. If you do it a lot then create a specialised class or function to encapsulate the logic and import it.

You can turn your code into:

params = dict()

if os.name == 'nt':
    startupinfo = subprocess.STARTUPINFO()
    startupinfo.dwFlags |= subprocess.STARTF_USESHOWWINDOW
    params['startupinfo'] = startupinfo

proc = subprocess.Popen(command, **params)

but that's not much better.

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