'type(v)' is model..instead of 'v is model' works,
in case not too late for someone else...
thanks Matt..
def get_admin(model):
for k,v in admin.site._registry.iteritems():
if type(v) is model:
return v
On Monday, January 11, 2010 at 9:37:47 PM UTC, Matt Schinckel wrote:
>
> On Jan 12, 4:11 am, Tomasz Zieliński
> <[email protected]> wrote:
> > On 11 Sty, 16:23, Marco Rogers <[email protected]> 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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