[pygtk] Accessing child widgets via instance variables
dvogel at intercarve.net
Sat Aug 30 01:00:25 WST 2008
If I inherit from gtk.Window as such:
class DocWindow(gtk.Window, libglade.GladeWrapper):
then program exits unexpectedly without any console output so I assumed
that pygtk did not allow this. Therefore, since my DocWindow class does
not inherit from gtk.Window, it has no show() method of it's own. If
this still isn't clear I will paste a test case.
Dominic Salemno wrote:
> If my_win is already an instance of GtkWindow then to show it you merely
> do something along this line: my_win.show()
> You should not have to do anything afterwards. If my_win does not
> represent your window, what is this representing? If something else
> entirely, choose a different variable name. Perhaps you could paste more
> code for us to see?
> Your variable, if it indeed defines a GtkWindow, should be declared as
> self.my_win = gtk.Window(gtk.WINDOW_TOPLEVEL)
> Even if my_win is instantiating a custom class which in turn is
> inheriting from gtk.Window, you would still perform the following to
> show the window: self.my_win.show()
> If this is not working, please paste more code.
> Sincerely, Dominic Salemno.
> Pádraig Brady wrote:
>> Drew Vogel wrote:
>>> Thanks, I have this mostly working. However it looks like instead of calling
>>> I now have to call
>>> Is this correct?
>> Yes. my_win.GtkWindow is just a shortcut I thought was useful.
>> To access other widgets, just use my_win.id.
>> pygtk mailing list pygtk at daa.com.au
>> Read the PyGTK FAQ: http://www.async.com.br/faq/pygtk/
> pygtk mailing list pygtk at daa.com.au
> Read the PyGTK FAQ: http://www.async.com.br/faq/pygtk/
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