[Fixed]-Return Custom 404 Error when resource not found in Django Rest Framework

by

22👍

✅

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.

👤Ernest

6👍

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')]
👤sabtikw

0👍

@Ernest Ten’s answer is correct. However, I would like to add some input if you’re using an application that handles both browser page loading and API as call.

I customized the custom404 function to this

def custom404(request, exception=None):
    requested_html = re.search(r'^text/html', request.META.get('HTTP_ACCEPT')) # this means requesting from browser to load html
    api = re.search(r'^/api', request.get_full_path()) # usually, API URLs tend to start with `/api`. Thus this extra check. You can remove this if you want.

    if requested_html and not api:
        return handler404(request, exception) # this will handle error in the default manner

    return JsonResponse({
        'detail': _('Requested API URL not found')
    }, status=404, safe=False)
👤Koushik Das

0👍

update of Ernest Ten answer:

You should also include
in urls.py

from django.http import JsonResponse
...other urls...
handler404 = lambda request, exception=None: JsonResponse({'error': '404: The resource was not found'}, status=404)
👤suhailvs

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