i know this isn't what you want, but I'd use the Content Type
middleware.
def getType(self):
if (self._type == None):
self._type = ContentType.objects.get_for_model(self)
return self._type
for me personally this worked out well, but mostly because I have an
abstract model that all my other models extend, so I had to define that
method, once....
On Wed, 2010-08-04 at 11:27 -0700, David.D wrote:
> I just wonder if there's some way requires writing nothing. Just like
> an attribute.
>
> thanks.
>
> On Aug 4, 11:53 pm, Scott Gould <zinck...@gmail.com> wrote:
> > Not that I'm aware of, in which case I'd say a tag (or even a filter)
> > *is* the "built-in way". No great hardship:
> >
> > @register.filter
> > def class_name(value):
> > return value.__class__.__name__
> >
> > On Aug 4, 11:13 am, "David.D" <dengyuanzh...@gmail.com> wrote:
> >
> >
> >
> > > There's no django's built-in way?
> >
> > > On Aug 4, 10:37 pm, Scott Gould <zinck...@gmail.com> wrote:
> >
> > > > How about writing a simple template tag that takes an object and
> > > > returns object.__class__.__name__?
> >
> > > > On Aug 4, 10:20 am, "David.D" <dengyuanzh...@gmail.com> wrote:
> >
> > > > > I did it by adding this to my models:
> > > > > def get_model_name(self):
> > > > > return self.__class__.__name__
> >
> > > > > And it works....
> >
> > > > > But I don't want to define the 'get_model_name' method in my model.
> >
> > > > > Is there a built-in way?
> >
> > > > > Thanks.
>
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