Return Custom 404 Error when resource not found in

问题:

I am learning Django Rest Framework, and also new to django. I want to return a custom 404 error in json when a client will access a resource which was not found.

My urls.py looks liks this:

urlpatterns = [
    url(r'^mailer/$', views.Mailer.as_view(), name='send-email-to-admin')
]

In which i have only one resource, which can be accessed through URI, http://localhost:8000/mailer/

Now, when a client access any other URI like http://localhost:8000/, API should return a 404-Not Found error like this:

{
    "status_code" : 404
    "error" : "The resource was not found"
}

Please suggest some answer with proper code snippets, if suitable.

回答1:

You are looking for handler404.

Here is my suggestion:

1. Create a view that should be called if none of the URL patterns match.
2. Add handler404 = path.to.your.view to your root URLconf.

Here is how it's done:

1. project.views

   from django.http import JsonResponse
   
   
   def custom404(request, exception=None):
       return JsonResponse({
           'status_code': 404,
           'error': 'The resource was not found'
       })
   

2. project.urls

   from project.views import custom404
   
   
   handler404 = custom404
   

Read error handling for more details.

Django REST framework exceptions may be useful as well.

回答2:

according to django documentation : Django runs through each URL pattern, in order, and stops at the first one that matches the requested URL. ref: https://docs.djangoproject.com/en/1.8/topics/http/urls/

so you can just add another url in urlpatterns after the one you created and it should match all url patterns and send them to a view that return the 404 code.

i.e :

urlpatterns = [
url(r'^mailer/$', views.Mailer.as_view(), name='send-email-to-admin'),
url(r'^.*/$',views.Error404.as_view(),name='error404')]