Open file in default app and exit in Windows

Tim Golden mail at
Wed Jan 28 17:07:04 CET 2015

On 28/01/2015 15:50, stephen.boulet at wrote:
> I am using the following to open a file in its default application in
> Windows 7:
> from subprocess import call
> filename = 'my file.csv' call('"%s"' % filename, shell=True)
> This still leaves a python process hanging around until the launched
> app is closed. Any idea how to get around?

This is somewhat clumsy. The built-in way is:

import os
os.startfile("my file.csv")


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