startfile() equivalent
Hye-Shik Chang
hyeshik at gmail.com
Wed Aug 18 19:19:08 PDT 2004
On Wed, 18 Aug 2004 11:09:43 +0100 (BST), Max Russell
<max_russell2000 at yahoo.co.uk> wrote:
> Hello-
>
> I'm scripting a little thing to pick a random MAME
> file and launch it (save me the hassle of choosing
> one).
>
> Does startfile work on BSD/Linux?
os.startfile() is only available in Windows.
> How can I do the equivalent to the python win32
> startfile()?
>
> If I cannot use this, how can I say execute this command?
>
It depends what your desktop environment is. If you run GNOME,
you can write it like this:
import os
def startfile(url):
os.system("gnome-open " + url) # quote if you can't trust user.
Hye-Shik
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