django return 404
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