How do you return 404 when resource is not found in Django REST Framework

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..

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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