当用户输入错误的url时，我的Django应用程序返回一个HTML错误.如何让DRF返回json格式的错误？
目前我的网址是
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),
]
但是如果用户转到127.0.0.1:8000/snip，他们得到的是html格式的错误，而不是json格式的错误.
简单地说，你可以用raise Http404，这是你的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)
你也可以用Response(status=status.HTTP_404_NOT_FOUND)来处理它，这个答案是如何处理它:https://stackoverflow.com/a/24420524/6396981
但之前，在你的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)
为了测试它，给出了一个使用curl命令的例子；
$ curl -X GET http://localhost:8000/snippets/<pk>/
# example;
$ curl -X GET http://localhost:8000/snippets/99999/
希望它能帮上忙.
如果您想用DRF处理所有错误404URL，DRF还提供了关于它的APIException，这个答案可能会对您有帮助；https://stackoverflow.com/a/30628065/6396981
我将举一个例子说明如何处理它；
1. 100
from rest_framework.exceptions import NotFound
def error404(request):
raise NotFound(detail="Error 404, page not found", code=404)
2. 100
from django.conf.urls import (
handler400, handler403, handler404, handler500)
from yourapp.views import error404
handler404 = error404
确保你的
DEBUG = False