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 。
答案 0 :(得分:7)
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']
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)
答案 1 :(得分:1)
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"})