Manually trigger Django email error report

Question

Django error reporting handles uncaught exceptions by sending an email, and (optionally) shows user a nice 500 error page.

This works very well, but in a few instances I'd like to allow users to continue with their business uninterrupted, but still have Django send me the email error report about the exception.

So basically: can I manually send email error report even if I catch the exception?

Of course, I'd like to avoid manually generating the error report email.

Answer 1

You can use the following code to send manually an email about a request and an exception e:

import sys import traceback from django.core import mail from django.views.debug import ExceptionReporter  def send_manually_exception_email(request, e):     exc_info = sys.exc_info()     reporter = ExceptionReporter(request, is_email=True, *exc_info)     subject = e.message.replace('\n', '\\n').replace('\r', '\\r')[:989]     message = "%s\n\n%s" % (         '\n'.join(traceback.format_exception(*exc_info)),         reporter.filter.get_request_repr(request)     )     mail.mail_admins(         subject, message, fail_silently=True,         html_message=reporter.get_traceback_html()     ) 

You can test it in a view like this:

def test_view(request):     try:         raise Exception     except Exception as e:         send_manually_exception_email(request, e) 

Answer 2

Just setup a simple log handler in your settings.

LOGGING = {     'version': 1,      'disable_existing_loggers': False,     'filters': {         'require_debug_false': {             '()': 'django.utils.log.RequireDebugFalse'         }     },     'handlers': {         'mail_admins': {             'level': 'ERROR',             'filters': ['require_debug_false'],             'class': 'django.utils.log.AdminEmailHandler'         },         'app': {             'level': 'ERROR',             'filters': ['require_debug_false'],             'class': 'django.utils.log.AdminEmailHandler'         },     },     'loggers': {         'django.request': {             'handlers': ['mail_admins'],             'level': 'ERROR',             'propagate': True,         },     } } 

and then in your view, you can do anything

 import logging  logger = logging.getLogger('app')   def some_view(request):      try:          # something          if something_wnet_wrong:              logger.error('Something went wrong!')          return some_http_response      except:          #something else          logger.error(sys.exc_info(), request)                  return some_other_response 

If you want detailed error report, you can try something like this.

You also need to take care of sensitive information.