Date: Thu, 19 Aug 2004 08:56:41 +0100 (BST) From: =?iso-8859-1?q?Max=20Russell?= <max_russell2000@yahoo.co.uk> To: Hye-Shik Chang <hyeshik@gmail.com> Cc: freebsd-python@freebsd.org Subject: Re: startfile() equivalent Message-ID: <20040819075641.97538.qmail@web25407.mail.ukl.yahoo.com> In-Reply-To: <4f0b69dc0408181919172d239d@mail.gmail.com>
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Is this soley for opening URLs though? I am wanting to
launch MAME with Python- could I substitute the URL
with the path to the application?
--- Hye-Shik Chang <hyeshik@gmail.com> wrote:
> On Wed, 18 Aug 2004 11:09:43 +0100 (BST), Max
> Russell
> <max_russell2000@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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