Manually trigger Django email error report

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)

Leave a Comment

Hata!: SQLSTATE[HY000] [1045] Access denied for user 'divattrend_liink'@'localhost' (using password: YES)