Question

I want to create a model that doesn’t map to a database table. Instead, stays in memory as a python object.

Actually, this model is supposed to represents normalised data from many other table-mapped models.

The other models store data which can be edited multiple times in a single day. Because of these multiple edits, I don’t want a table-mapped-model that performs normalisations/calculations and stores them in a database as, this stored data can go out of date right away.

Every time this normalised model is accessed (via admin), I want it to perform the normalisations on data from the other models from scratch (So that it can show the most up to date data) and behave just like a normal model would under admin like Showing the list view and a detailed view for each row.

Edit after Shintoist’s answer:

@Shintoist Thanks for the clearing things out and providing a usable approach. I have just implemented it but hitting a small wall in the end 🙂

@skirmantas: Yes, the calculations are in a separate object. This object is being passed into the custom views.

Problem: One problem is that under admin.py, I have created an modeladminclass for this object(which doesn’t inherit models.Model) so my custom views can overide the changelist view and changeview.
I then use admin.site.register() to register this model-like class and the modeladmin. But, since this model is not a django model at all (as it is an independant python object in memory) admin.site.register() throws a ” ‘type’ object is not iterable” error. I don’t want to use the url.py instead of admin.py as it’s meant for the frontend while Im trying to overide the backend-admin.

Answer 1

hmmm. Thanks for your help everyone. The solution I have come up (with your help ofcourse) is as follows:

I have two custom templates:

   my_model_list.html
   my_model_detail.html

Under views.py:

class MyModel(object):
    # ... Access other models
    # ... process / normalise data 
    # ... store data

@staff_member_required
def my_model_list_view(request) #show list of all objects
    #. . . create objects of MyModel . . .
    #. . . call their processing methods . . .
    #. . . store in context variable . . . 
    r = render_to_response('admin/myapp/my_model_list.html', context, RequestContext(request))
    return HttpResponse(r)

@staff_member_required
def my_model_detail_view(request, row_id) # Shows one row (all values in the object) in detail     
    #. . . create object of MyModel . . .
    #. . . call it's methods . . .
    #. . . store in context variable . . . 
    r = render_to_response('admin/myapp/my_model_detail.html', context, RequestContext(request))
    return HttpResponse(r)

Under the main django urls.py:

urlpatterns = patterns( 
    '',
    (r'^admin/myapp/mymodel/$', my_model_list_view),
    (r'^admin/myapp/mymodel/(\d+)/$', my_model_detail_view),
    ( r'^admin/', include( admin.site.urls ) )
)

As you’ve already noticed, I had to insert url patterns to my url.py file. I don’t know if thats the best way to do it, as I reckon, the url.py file is not meant for admin related pages. It’s only for the site frontend.