os.startfile: Why is there no arguments option?

Wed Oct 12 08:18:00 EDT 2011

On Oct 12, 11:27 am, Thomas Rachel <nutznetz-0c1b6768-bfa9-48d5-
a470-7603bd3aa... at spamschutz.glglgl.de> wrote:
> Am 12.10.2011 10:22 schrieb Christian Wutte:
>> > Hello all,
> > as stated in the docs [1] os.startfile relies on Win32 ShellExecute().
> > So maybe someone can explain it to me, why there is no support for
> > program arguments.
>> Because it is intended to start an arbitrary file of any type (.txt,
> .doc, ...) For this operations, there is no parameter support.
>
Yes, but isn't that also the case for ShellExecute()? In the MSDN page
[1] for the parameters can be read:
"If lpFile specifies an executable file, this parameter is a pointer
to a null-terminated string that specifies the parameters to be
passed to the application. The format of this string is determined by
the verb that is to be invoked. If lpFile specifies a document file,
lpParameters should be NULL."
So the parameter should be optional for sure.
> > That's quite a pity since os.startfile is the easiest way for an
> > elevated run (with 'runas' as option)
>> Obviously not.
>>  > and without arguments of limited use.
>> So it isn't the asiest way.
>> Have you tried os.system() and/or subprocess.Popen() resp. .call()?
>
Yes. I should have noticed, that in my case I want to make a manual
elevated call. With programs that need elevated rights in the first
place os.start() or subprocess.Popen(.. , shell=True) is fine.
Chrisitan