Open a file with default handler app?

[andrei.avk]

Hi, I searched for this on google and in this group, but my awesome
google-fu powers failed me. Is there a way to open any file using
default program that'd open it? In other words, to do the same action
as double-clicking in windows explorer? And secondly, is there a way
to do the same thing for linux that'd work across all desktop
environments and distributions, or at least in all major ones? What
I'm trying to do here is to have records (or items) in my app where
you could attach any kind of file and open it from there by clicking
'open'. Then it would go and do something like os.system("launch %s" %
filename). So any way to do this except for keeping your own
dictionary of file types and relevant apps? thx, -ak

[Wubbulous]

On Mar 11, 9:24*pm, [e-mail] wrote:

Hi, I searched for this on google and in this group, but my awesome
google-fu powers failed me. Is there a way to open any file using
default program that'd open it? In other words, to do the same action
as double-clicking in windows explorer? And secondly, is there a way
to do the same thing for linux that'd work across all desktop
environments and distributions, or at least in all major ones? What
I'm trying to do here is to have records (or items) in my app where
you could attach any kind of file and open it from there by clicking
'open'. Then it would go and do something like os.system("launch %s" %
filename). So any way to do this except for keeping your own
dictionary of file types and relevant apps? thx, -ak

Hey there,
I've had to do the same things for a program that I'm writing. The
following command should do the trick:

os.startfile("yourfilehere")

from the os module. Hope this helps!

[andrei.avk]

On Mar 11, 11:50*pm, [e-mail] wrote:

>
Hey there,
I've had to do the same things for a program that I'm writing. The
following command should do the trick:

os.startfile("yourfilehere")

from the os module. Hope this helps!

That's perfect, thanks a ton++!