Question

我有一个基于类的通用视图：

class ProjectDetails(mixins.RetrieveModelMixin,
                     mixins.UpdateModelMixin,
                     generics.GenericAPIView):
    queryset = Project.objects.all()
    # Rest of definition

在我的urls.py中，我有：

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

使用不存在的ID调用API时，它将返回HTTP 404响应，内容如下：

{
    "detail": "Not found."
}

是否可以修改此响应？

我只需要为此视图自定义错误消息。

Answer 1

此解决方案会影响所有视图：

您当然可以提供自定义异常处理程序：Custom exception handling

from rest_framework.views import exception_handler
from rest_framework import status

def custom_exception_handler(exc, context):
    # Call REST framework's default exception handler first,
    # to get the standard error response.
    response = exception_handler(exc, context)

    # Now add the HTTP status code to the response.
    if response.status_code == status.HTTP_404_NOT_FOUND:
        response.data['custom_field'] = 'some_custom_value'

    return response

确定可以跳过默认的rest_framework.views.exception_handler并将其完全设为原始。

注意：切记在django.conf.settings.REST_FRAMEWORK['EXCEPTION_HANDLER']

中提及您的处理程序

特定视图的解决方案：

from rest_framework.response import Response
# rest of the imports

class ProjectDetails(mixins.RetrieveModelMixin,
                     mixins.UpdateModelMixin,
                     generics.GenericAPIView):
    queryset = Project.objects.all()

    def handle_exception(self, exc):
        if isinstance(exc, Http404):
            return Response({'data': 'your custom response'}, 
                            status=status.HTTP_404_NOT_FOUND)

        return super(ProjectDetails, self).handle_exception(exc)

Answer 2

可以通过将update，retrieve之类的特定方法重写为：

from django.http import Http404
from rest_framework.response import Response


class ProjectDetails(mixins.RetrieveModelMixin,
                     mixins.UpdateModelMixin,
                     generics.GenericAPIView):
    queryset = Project.objects.all()

    def retrieve(self, request, *args, **kwargs):
        try:
            return super().retrieve(request, *args, **kwargs)
        except Http404:
            return Response(data={"cusom": "message"})

    def update(self, request, *args, **kwargs):
        try:
            return super().update(request, *args, **kwargs)
        except Http404:
            return Response(data={"cusom": "message"})

404错误的Django Rest Framework自定义消息

2 个答案: