python - 404 错误的 Django Rest Framework 自定义消息

Question

我有一个基于类的通用 View :

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())
]

当使用不存在的 ID 调用 API 时，它会返回 HTTP 404 响应，其内容为:

{
    "detail": "Not found."
}

是否可以修改此响应？

我需要仅为此 View 自定义错误消息。

Answer 1

此解决方案影响所有 View :

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

当然你可以跳过默认的 rest_framework.views.exception_handler 并使其完全原始。

注意:记得在 django.conf.settings.REST_FRAMEWORK['EXCEPTION_HANDLER'] 中提及您的处理程序

针对特定 View 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)