Sick Monkey escribió: > I ran into another slight problem. And I attempted to fix it, but have > not been able to do so yet. If a filename does not contain a space, > then this method works like a charm. But if there is a space then the > code throws a nasty error.
Ok, the issue is that subprocess.Popen needs a list of the arguments. So, for simplicity, I made an string and then splitted it: >>> s = "ls -l /etc" >>> s.split() ['ls', '-l', '/etc'] >>> This does not work if you have spaces, even if you escape them: >>> s = "ls -l '/etc/mi file'" >>> s.split() ['ls', '-l', "'/etc/mi", "file'"] >>> But you can always build the list by hand: >>> subprocess.Popen(["ls", "-l", "'/etc/mi file'"], ... Regards, -- . Facundo . Blog: http://www.taniquetil.com.ar/plog/ PyAr: http://www.python.org/ar/ -- http://mail.python.org/mailman/listinfo/python-list
