Django snippets: Simple Exception Response for AJAX debugging

from django.conf import settings
from django.http import HttpResponseServerError

class AJAXSimpleExceptionResponse:
    def process_exception(self, request, exception):
        if settings.DEBUG:
            if request.META.get('HTTP_X_REQUESTED_WITH', None) == 'XMLHttpRequest':
                import sys, traceback
                (exc_type, exc_info, tb) = sys.exc_info()
                response = "%s\n" % exc_type.__name__
                response += "%s\n\n" % exc_info
                response += "TRACEBACK:\n"    
                for tb in traceback.format_tb(tb):
                    response += "%s\n" % tb
                return HttpResponseServerError(response)

When debugging AJAX with Firebug, if a response is 500, it is a pain to have to view the entire source of the pretty exception page. This is a simple middleware that just returns the exception without any markup. You can add this anywhere in your python path and then put it in you settings file. It will only unprettify your exceptions when there is a XMLHttpRequest header. Tested in FF2 with the YUI XHR. Comments welcome.

Author:: newmaniese
Posted:: March 19, 2008
Language:: Python
Tags:: ajax exception-handling middleware
Score:: 11 (after 11 ratings)

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luftyluft (on March 19, 2008):

Nice idea - thanks!

skabber (on March 19, 2008):

This will be great. Although I have gotten good at finding the exception in the source of the pretty error pages :)

robhudson (on March 20, 2008):

Works great. Thanks!

trevor (on March 23, 2008):

You can also document.write() on 500 in the javascript to display the error instead of trying to read it in firebug.

kioopi (on April 15, 2008):

I use it, too.

I use the Poster firefox add-on for a lot of Ajax-Debugging, but it seems buggy for HTTP_ACCEPT headers.

Thanks.

Django snippets

Simple Exception Response for AJAX debugging

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