[Solved]-Can django-tastypie display a different set of fields in the list and detail views of a single resource?



You would have to specify all fields in the actual ModelResource then override the get_list method to filter out only the fields you want to show. See the internal implementation of get_list on Resource to see how to override it.

However, note this will only apply on GET requests, you should still be able to POST/PUT/PATCH on the resource with all fields if you authorization limits allow you to do so.

In a nut shell, you want to hot patch the internal field list before full_dehydrate is called on all ORM objects returned by obj_get_list.

Alternatively, you can let the full dehydrate mechanism take place and just at the end of it remove the fields you don’t want to show if you don’t care about squeezing out as much as speed as possible. Of course you would need to do this only if the URL is invoked as a consequence of get_list call. There is a convenience method for this alter_list_data_to_serialize(request, to_be_serialized).

Just do:

class SomeResource(Resource):
    class Meta(...):
         field_list_to_remove = [ 'field1', 'field2' ]

    def alter_list_data_to_serialize(request, to_be_serialized):
        for obj in to_be_serialized['objects']:
            for field_name in self._meta.field_list_to_remove:
                del obj.data[field_name]
        return to_be_serialized


You can also now use the use_in attribute on a field to specify the relevant resource to show the field in. This can either be list or detail, or a callback.

πŸ‘€Ben Eliott


There is an open issue for this on GitHub, with a number of workarounds suggested there.



Can also use the dehydrate(self, bundle) method.

def dehydrate(self, bundle):
del bundle.data['attr-to-del]
return bundle

πŸ‘€Ronn Macc

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