problems with opening files due to file's path

Alexnb alexnbryan at gmail.com
Wed Jun 11 05:07:25 CEST 2008


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 <alexnbryan at gmail.com> 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
> 
> 

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