Djrill: Mandrill Transactional Email for Django

Djrill integrates the Mandrill transactional email service into Django.

In general, Djrill "just works" with Django's built-in django.core.mail package. It includes:

• Support for HTML, attachments, extra headers, and other features of Django's built-in email
• Mandrill-specific extensions like tags, metadata, tracking, and MailChimp templates
• Optional support for Mandrill inbound email and other webhook notifications, via Django signals
• An optional Django admin interface

Djrill is released under the BSD license. It is tested against Django 1.3---1.6 (including Python 3 support with Django 1.5+).

Resources

• Full documentation: https://djrill.readthedocs.org/en/latest/
• Package on PyPI: https://pypi.python.org/pypi/djrill
• Project on Github: https://github.com/brack3t/Djrill

Djrill 1-2-3

1. Install Djrill from PyPI:

   $ pip install djrill

2. Edit your project's settings.py:

   INSTALLED_APPS = (
       ...
       "djrill"
   )
   
   MANDRILL_API_KEY = "<your Mandrill key>"
   EMAIL_BACKEND = "djrill.mail.backends.djrill.DjrillBackend"

3. Now the regular Django email functions will send through Mandrill:

   from django.core.mail import send_mail
   
   send_mail("It works!", "This will get sent through Mandrill",
       "Djrill Sender <djrill@example.com>", ["to@example.com"])

   You could send an HTML message, complete with custom Mandrill tags and metadata:

   from django.core.mail import EmailMultiAlternatives
   
   msg = EmailMultiAlternatives(
       subject="Djrill Message",
       body="This is the text email body",
       from_email="Djrill Sender <djrill@example.com>",
       to=["Recipient One <someone@example.com>", "another.person@example.com"],
       headers={'Reply-To': "Service <support@example.com>"} # optional extra headers
   )
   msg.attach_alternative("<p>This is the HTML email body</p>", "text/html")
   
   # Optional Mandrill-specific extensions:
   msg.tags = ["one tag", "two tag", "red tag", "blue tag"]
   msg.metadata = {'user_id': "8675309"}
   
   # Send it:
   msg.send()

See the full documentation for more features and options.