[pygtk] Ok. A new try...

[Gonzalo Aguilar Delgado]

Hi Tim!

It's better avoid recursion if you have too much deep levels.
Will try this also...

Thank you for your help.
This will be keept here and will help a lot of people also.

I'm new to python (only 've done few things in blender) but loveling
it....



Avoid recursion

> John Ehresman wrote:
> > Gonzalo Aguilar Delgado wrote:
> >> How do I find a widget (say a GtkEntry) inside it with the name
> >> "txtbox"?
> > 
> > You need to go through all the children and find the one with the name. 
> >   Try the following:
> > 
> > def all_descendants(w, flags=None):
> >   """ Generator to traverse all descendants of a widget in depth-first 
> >   order.  Assumes that the widget hierarchy doesn't change during the 
> >   traversal.  Use flags to only return widgets where widget & flags is
> >   true.  If a widget is filtered out, all of its descendants are also
> >   filtered out. """
> > 
> >   if not isinstance(w, gtk.Container):
> >     return
> > 
> >   for child in w.get_children():
> >     if flags is None or child.flags() & flags:
> >       yield child
> > 
> >       for g_kid in all_descendants(child):
> >         yield g_kid
> > 
> > def find_named(parent, name):
> >   for w in all_descendants(parent):
> >     if w.get_name() = name:
> >       return w
> >   return None
> > 
> > entry = find_named(frame, 'txtbox')
> > 
> > You probably do want to make sure that only one widget with the name 
> > 'txtbox' is a descendant of frame.  gtk does not enforce this for you.
> 
> I generally prefer to avoid the recursion and use something like this:
> 
>      def all_descendants(root):
>          stack = [root]
>          while stack:
>              w = stack.pop()
>              yield w
>              if isinstance(w, gtk.Container):
>                  stack.extend(reversed(w.get_children()))
>              if isinstance(w, gtk.MenuItem):
>                  submenu = w.get_submenu()
>                  if submenu:
>                      stack.append(submenu)
> 
> Submenus don't count as children but for most things it's useful to have
> them listed in the same way.  I would also recommend putting all the
> widgets into a dictionary if you'll be doing these lookups often:
> 
>      def named_widgets(root):
>           r = {}
>           for w in all_descendants(root):
>               name = w.get_name()
>               if name:
>                   r[name] = w
>           return r
> 
-- 
Gonzalo Aguilar Delgado - Ingeniero en Informática
        [ Seguridad & Medios de pago ]