How do you return 404 when resource is not found i

Question

When a user inputs a url that is wrong, my Django app returns an HTML error. How can I get DRF to return a json formatted error?

Currently my urls is

from django.conf.urls import url
from snippets import views

urlpatterns = [
    url(r'^snippets/$', views.snippet_list),
    url(r'^snippets/(?P<pk>[0-9]+)/$', views.snippet_detail),
]

but if a user goes to 127.0.0.1:8000/snip They get the html formatted error rather than a json formatted error.

Answer 1

Simply way to do it, you can use raise Http404, here is your views.py

from django.http import Http404

from rest_framework import status
from rest_framework.response import Response
from rest_framework.views import APIView

from yourapp.models import Snippet
from yourapp.serializer import SnippetSerializer


class SnippetDetailView(APIView):

    def get_object(self, pk):
        try:
            return Snippet.objects.get(pk=pk)
        except Snippet.DoesNotExist:
            raise Http404

    def get(self, request, pk, format=None):
        snippet = self.get_object(pk)
        serializer = SnippetSerializer(snippet)
        return Response(serializer.data, status=status.HTTP_200_OK)

You also can handle it with Response(status=status.HTTP_404_NOT_FOUND), this answer is how to do with it: https://stackoverflow.com/a/24420524/6396981

But previously, inside your serializer.py

from rest_framework import serializers

from yourapp.models import Snippet


class SnippetSerializer(serializers.ModelSerializer):
    user = serializers.CharField(
        source='user.pk',
        read_only=True
    )
    photo = serializers.ImageField(
        max_length=None,
        use_url=True
    )
    ....

    class Meta:
        model = Snippet
        fields = ('user', 'title', 'photo', 'description')

    def create(self, validated_data):
        return Snippet.objects.create(**validated_data)

To test it, an example using curl command;

$ curl -X GET http://localhost:8000/snippets/<pk>/

# example;

$ curl -X GET http://localhost:8000/snippets/99999/

Hope it can help..

---

Update

If you want to handle for all error 404 urls with DRF, DRF also provide about it with APIException, this answer may help you; https://stackoverflow.com/a/30628065/6396981

I'll give an example how do with it;

1. views.py

from rest_framework.exceptions import NotFound

def error404(request):
    raise NotFound(detail="Error 404, page not found", code=404)

2. urls.py

from django.conf.urls import (
  handler400, handler403, handler404, handler500)

from yourapp.views import error404

handler404 = error404

Makesure your DEBUG = False