IMHO, the fact that there is no way to wait for the application to
finish is major deficiency with os.startfile() -- and why I often
cannot use it instead of other techniques [such as the much more
involved but limited os.spawnv() function].
I don't if if this is just a limitation of the implementation in the
Python os module or one with the underlying OS (Windoze) -- I suspect
the latter.
Regards,
-Martin
Dennis Lee Bieber wrote:
> On Thu, 22 Sep 2005 15:16:09 +0100, Dan <[EMAIL PROTECTED]> declaimed
> the following in comp.lang.python:
>
> > > I would like to know how to open a PDF document from a python script
> >
> > You mean open it and display it to the user? Under Windows you may be
> > able to get away with just "executing" the file (as though it were an
> > executable):
> >
> > import os
> > os.system("c:/path/to/file.pdf")
>
> For Windows, os.startfile() may be better...
>
> >>> help(os.startfile)
> Help on built-in function startfile:
>
> startfile(...)
> startfile(filepath) - Start a file with its associated application.
>
> This acts like double-clicking the file in Explorer, or giving the
> file
> name as an argument to the DOS "start" command: the file is opened
> with whatever application (if any) its extension is associated.
>
> startfile returns as soon as the associated application is launched.
> There is no option to wait for the application to close, and no way
> to retrieve the application's exit status.
>
> The filepath is relative to the current directory. If you want to
> use
> an absolute path, make sure the first character is not a slash
> ("/");
> the underlying Win32 ShellExecute function doesn't work if it is.
>
> >>>
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