Question

Reader s Digest version: How do I get data (either single row if specified or full table) from a model by taking a URL argument of the model s name without hardcoding the URLconfs to match each model? For more details, read on:

I m making an app that has three models, and I want to make a simple view that takes a name of a model, and spit out the model s default manager s Model.manager.all(), and if there s a slug, a single object matching that slug. I don t know how to do this, so I m running into problems with having to make views/URLconfs for EACH model individually.

Here s what the URLconfs are looking like:

(r ^model1/$ ,  model1_index_view ,  model1_index ),
(r ^model1/(?P<slug>[-w]+)/$ ,  model1_detail_view ,  model1_detail ),
(r ^model2/$ ,  model2_index_view ,  model2_index ),

It goes on a little bit longer, but I think you get the picture. I m ending up hardcoding a relatively large amount of URLconfs to do something that I m thinking I can do with perhaps one View that takes a Model name as an argument and optionally a slug. My concern is, what happens if someone specifies a Model name of say... User? Is there a snippet of code to get a list of models from an app and make sure it matches one of em and not a model from contrib.auth or another app?

Answer 1

Use get_model:

from django.db.models import get_model

def my_view(request, model_name, item_slug):
    try:
        model = get_model( app_name , model_name)
    except:
        ## throw an error
        pass
    objects = model.objects.get(slug=item_slug)

Then in urls:

 (r ^(?P<model_name>[-w]+/(?P<slug>[-w]+)/$ ,  model_detail_view ,  model_detail ),

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