On Tue, Jul 15, 2008 at 1:22 PM, Torsten Bronger
<[EMAIL PROTECTED]> wrote:
> Arien writes:
>> You'll have to make layer._meta.verbose_name available to the
>> template under some other name.
>
> Unfortunately, I use model polymorphism and model inheritance in
> this case and can't pass anything but the common base class to the
> template.
How about sticking this on the base class then?
@property
def verbose_name(self):
return self._meta.verbose_name
> Therefore, I collect all verbose_names in an "all_labels" dict in my
> models.py:
>
> all_labels = {}
> for cls in [cls for cls in _globals.values() if inspect.isclass(cls) and
> issubclass(cls, models.Model)]:
> local_labels = {}
> for field in cls._meta.local_fields:
> local_labels[field.name] = field.verbose_name
> all_labels[cls.__name__] = local_labels
Oh, I see, you want the verbose_name of each field. I suppose you
could write a templatetag that you could use like this:
<td>{% verbose_name layer.heating_temperature %}</td>
<td>{{ layer.heating_temperature }}</td>
After parsing the templatetag's argument, you'd resolve the model
instance and insert this in the template:
model_instance._meta.get_field(field_name).verbose_name
Arien
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"Django users" group.
To post to this group, send email to django-users@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at
http://groups.google.com/group/django-users?hl=en
-~----------~----~----~----~------~----~------~--~---
