Re: Open file in default app and exit in Windows

2015-01-28 Thread Tim Chase
On 2015-01-28 07:50, stephen.bou...@gmail.com 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) You can try import os filename = 'my file.csv' os.st

Re: Open file in default app and exit in Windows

2015-01-28 Thread stephen . boulet
On Wednesday, January 28, 2015 at 10:07:25 AM UTC-6, Tim Golden wrote: > On 28/01/2015 15:50, stephen...@gmail.com wrote 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"'

Re: Open file in default app and exit in Windows

2015-01-28 Thread Tim Golden
On 28/01/2015 15:50, stephen.bou...@gmail.com 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 t