That would work, but not for what I want. See the file could be anywhere on the user's system and so the entire path will be unique, and that didn't work with a unique path. What is the subprocess module you are talking about?
Mike Driscoll wrote: > > On Jun 10, 1:25 pm, "Thomas Morton" <[EMAIL PROTECTED]> > wrote: >> maybe try string substitution... not sure if that's really the BEST way >> to >> do it but it should work >> >> startfile(r"%s"%variable) > > > I concur. That should work. A slightly more in depth example (assuming > Windows): > > os.startfile(r'C:\Documents and Settings\%s\Desktop\myApp.exe' % > username) > > or > > os.startfile(r'C:\Program Files\%s' % myApp) > > Hopefully this is what you are talking about. If you were referring to > passing in arguments, than you'll want to use the subprocess module > instead. > > >> >> -------------------------------------------------- >> From: "Alexnb" <[EMAIL PROTECTED]> >> Sent: Tuesday, June 10, 2008 7:05 PM >> To: <[EMAIL PROTECTED]> >> Subject: Re: problems with opening files due to file's path >> >> >> >> > Well, now i've hit another problem, this time being that the path will >> be >> > a >> > variable, and I can't figure out how to make startfile() make it raw >> with >> > a >> > variable, if I put startfile(r variable), it doesn't work and >> > startfile(rvariable) obviously won't work, do you know how to make that >> > work >> > or better yet, how to take a regular string that is given and make >> every >> > single "\" into a double "\\"? >> > > <snip> > > Mike > -- > http://mail.python.org/mailman/listinfo/python-list > > -- View this message in context: http://www.nabble.com/problems-with-opening-files-due-to-file%27s-path-tp17759531p17762276.html Sent from the Python - python-list mailing list archive at Nabble.com. -- http://mail.python.org/mailman/listinfo/python-list
