On Oct 12, 11:27 am, Thomas Rachel <nutznetz-0c1b6768-bfa9-48d5- a470-7603bd3aa...@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 -- http://mail.python.org/mailman/listinfo/python-list
