Modified exception handler to throw django error page in case of 500 error (#4172)

Show Traceback HTML in browsable API
This commit is contained in:
Akhil Lawrence 2016-08-10 19:54:32 +05:30 committed by Tom Christie
parent 378b04eeaa
commit fa4ce50be7

View File

@ -3,13 +3,17 @@ Provides an APIView class that is the base of all views in REST framework.
""" """
from __future__ import unicode_literals from __future__ import unicode_literals
import sys
from django.conf import settings
from django.core.exceptions import PermissionDenied from django.core.exceptions import PermissionDenied
from django.db import models from django.db import models
from django.http import Http404 from django.http import Http404
from django.http.response import HttpResponseBase from django.http.response import HttpResponse, HttpResponseBase
from django.utils import six from django.utils import six
from django.utils.encoding import smart_text from django.utils.encoding import smart_text
from django.utils.translation import ugettext_lazy as _ from django.utils.translation import ugettext_lazy as _
from django.views import debug
from django.views.decorators.csrf import csrf_exempt from django.views.decorators.csrf import csrf_exempt
from django.views.generic import View from django.views.generic import View
@ -91,7 +95,11 @@ def exception_handler(exc, context):
set_rollback() set_rollback()
return Response(data, status=status.HTTP_403_FORBIDDEN) return Response(data, status=status.HTTP_403_FORBIDDEN)
# Note: Unhandled exceptions will raise a 500 error. # throw django's error page if debug is True
if settings.DEBUG:
exception_reporter = debug.ExceptionReporter(context.get('request'), *sys.exc_info())
return HttpResponse(exception_reporter.get_traceback_html(), status=500)
return None return None