I don't think you understand it doesn't matter how the variable gets there, the same code is run regardless, I have no problem with the GUI, but you asked, and so I told you. the code os.startfile(.... is run if there is a GUI or it is a console app.
Carsten Haese-2 wrote: > > Alexnb wrote: >> Okay, I don't understand how it is too vague, but here: >> > > [snip a bunch of irrelevant examples...] >> >> Did I clarify? > > No. Earlier you wrote: > >>> On 2008-06-11, Alexnb <[EMAIL PROTECTED]> wrote: >>>> I am using GUI, Tkinter to be exact. But regardless of how the >>>> path gets there, it needs to opened correctly. > > This implies that the file doesn't get opened correctly if the file name > is entered/chosen in the GUI. Yet, the examples you posted don't contain > any GUI code whatsoever. They merely demonstrate that you don't have a > firm grasp on how backslashes in string literals are treated. > > So, this begs the question, do you actually have any GUI code that is > failing, or are you just worried, given the problems you had with string > literals, that the GUI code you have yet to write will fail in the same > way? > > If this is the case, you should just write the GUI code and try it. It > might just work. Backslashes entered into a GUI text box are not treated > the same as backslashes in a Python string literal. > > If, on the other hand, you do have some GUI code for getting the file > name from the user, and that code is failing, then please, show us THAT > CODE and show us how it's failing. > > -- > Carsten Haese > http://informixdb.sourceforge.net > -- > 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-tp17759531p17769178.html Sent from the Python - python-list mailing list archive at Nabble.com. -- http://mail.python.org/mailman/listinfo/python-list
