I'm currently using the site._registry approach. I was hoping there
was either a more "official" way.
Incidentally, the registry is a dict so you should be able to access
the model directly.
from django.contrib import admin
def get_admin(model):
if model in admin.site._registry
return admin.site._registry[model]
return None
I haven't testing this because I'm doing something slightly different
:Marco
On Jan 11, 4:37 pm, Matt Schinckel <matt.schinc...@gmail.com> wrote:
> On Jan 12, 4:11 am, Tomasz Zieliñski
>
>
>
> <tomasz.zielin...@pyconsultant.eu> wrote:
> > On 11 Sty, 16:23, Marco Rogers <marco.rog...@gmail.com> wrote:
>
> > > I'm reposting this from earlier to see if I have better luck. I need
> > > to be able to get the ModelAdmin associated with a model at runtime.
> > > Similar to how django.db.models.get_model allows you to retrieve a
> > > model.
>
> > > admin_class = get_admin(Model)
>
> > > Is this possible?
>
> > > The original post has more info.
>
> > >http://groups.google.com/group/django-users/browse_thread/thread/0cd0...
>
> > Maybe something like this:
>
> > 1. Write something like django.contrib.admin.autodiscover, to get all
> > ModelAdmins
> > 2. Iterate those ModelAdmins, retrieve models they are attached to and
> > store this mapping in a dict.
> > 3. def get_admin(model):
> > return admin_to_model[model]
>
> You should be able to iterate through admin.site._registry, and pull
> out the one you need.
>
> from django.contrib import admin
> def get_admin(model):
> for k,v in admin.site._registry.iteritems():
> if v is model:
> return v
>
> Written in a browser.
>
> Matt.
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