Django REST Framework: Exclude a specific route while retain standard ViewSets

Question

I am using Django REST Framework (DRF) for my API. In particular, I have set up several routes using the ModelViewSet, without much additional coding. In particular, I have a route like

/api/v1/collection/<key>

From all the possible keys, I do need to block one specific key, say special. Any request with any HTTP verb to /api/v1/collection/special should result in a HTTP 404 error.

I can think of two approaches:

Override some method on the ViewSet hierarchy. Which one?
In urls.py, set up a higher-precedence URL route that intercepts this URL, like path("collection/special", view=<404View>, name="collection-exception"). Does this make sense? What would be the appropriate exception view to route to?

What is the recommended approach? Any of the above, or something else?

score 0 · Answer 1 · answered Nov 03 '20 at 16:25

0

If you are using the ModelViewSet then you should overwrite the get_queryset() method. If key value is wrong then raise the Exception otherwise return the queryset:

from rest_framework.exceptions import NotFound 

class MyModelViewSet(viewsets.ModelViewSet):
    # ...
    def get_queryset(self):
        key = self.kwargs.get("key")
        if key == "special":
            raise NotFound()
        return MyModel.objecs.filter(pk=key)

answered Nov 03 '20 at 16:25

pplonski

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Does this also work for `HEAD` and `OPTIONS` requests? – Ulrich Schuster Nov 03 '20 at 16:30
I think HEAD and OPTIONS are not used to get the response. They have different purposes. The 404 should be returned for GET. – pplonski Nov 03 '20 at 17:45
Ah, I see that my question was imprecise. I would like to "hide" the resource with the `special` key. Hence, all HTTP verbs should fail. – Ulrich Schuster Nov 03 '20 at 18:12
Try to add a URL that will raise an exception or show an error view when called – pplonski Nov 04 '20 at 09:11

Django REST Framework: Exclude a specific route while retain standard ViewSets

1 Answers1