This package implements an email backend for Django that relies on sendgrid's REST API for message delivery.
It is under active development, and pull requests are more than welcome!
To use the backend, simply install the package (using pip), set the EMAIL_BACKEND
setting in Django, and add a SENDGRID_API_KEY
key (set to the appropriate value) to your Django settings.
pip install django-sendgrid-v5
EMAIL_BACKEND = "sendgrid_backend.SendgridBackend"
SENDGRID_API_KEY = os.environ["SENDGRID_API_KEY"]
SENDGRID_SANDBOX_MODE_IN_DEBUG = True/False
.
SENDGRID_ECHO_TO_STDOUT
will echo to stdout or any other file-like
object that is passed to the backend via the stream
kwarg.SENDGRID_TRACK_EMAIL_OPENS
- defaults to true and tracks email open events via the Sendgrid service. These events are logged in the Statistics UI, Email Activity interface, and are reported by the Event Webhook.SENDGRID_TRACK_CLICKS_HTML
- defaults to true and, if enabled in your Sendgrid account, will tracks click events on links found in the HTML message sent.SENDGRID_TRACK_CLICKS_PLAIN
- defaults to true and, if enabled in your Sendgrid account, will tracks click events on links found in the plain text message sent.SENDGRID_HOST_URL
- Allows changing the base API URI. Set to https://api.eu.sendgrid.com
to use the EU region.from django.core.mail import send_mail
send_mail(
'Subject here',
'Here is the message.',
'from@example.com',
['to@example.com'],
fail_silently=False,
)
First, create a dynamic template and copy the ID.
from django.core.mail import EmailMessage
msg = EmailMessage(
from_email='to@example.com',
to=['to@example.com'],
)
msg.template_id = "your-dynamic-template-id"
msg.dynamic_template_data = {
"title": foo
}
msg.send(fail_silently=False)
from django.core.mail import EmailMessage
msg = EmailMessage(
from_email='to@example.com',
to=['to@example.com'],
cc=['cc@example.com'],
bcc=['bcc@example.com'],
)
# Personalization custom args
# https://sendgrid.com/docs/for-developers/sending-email/personalizations/
msg.custom_args = {'arg1': 'value1', 'arg2': 'value2'}
# Reply to email address (sendgrid only supports 1 reply-to email address)
msg.reply_to = 'reply-to@example.com'
# Send at (accepts an integer per the sendgrid docs)
# https://docs.sendgrid.com/for-developers/sending-email/scheduling-parameters#send-at
msg.send_at = 1600188812
# Transactional templates
# https://sendgrid.com/docs/ui/sending-email/how-to-send-an-email-with-dynamic-transactional-templates/
msg.template_id = "your-dynamic-template-id"
msg.dynamic_template_data = { # Sendgrid v6+ only
"title": foo
}
msg.substitutions = {
"title": bar
}
# Unsubscribe groups
# https://sendgrid.com/docs/ui/sending-email/unsubscribe-groups/
msg.asm = {'group_id': 123, 'groups_to_display': ['group1', 'group2']}
# Categories
# https://sendgrid.com/docs/glossary/categories/
msg.categories = ['category1', 'category2']
# IP Pools
# https://sendgrid.com/docs/ui/account-and-settings/ip-pools/
msg.ip_pool_name = 'my-ip-pool'
msg.send(fail_silently=False)
How to change a Sender's Name ?
from_email="John Smith <john.smith@example.org>"
You can just include the name in the fromemail field of the EmailMessage
_ class
msg = EmailMessage(
from_email='Sender Name <from@example.com>',
to=['to@example.com'],
)
How to make mails to multiple users private (hide all the email addresses to which the mail is sent) to each person (personalization) ?
Setting the make_private
attribute to True
will help us achieve it
msg = EmailMessage(
from_email='Sender Name <from@example.com>',
to=['to@example.com','abc@example.com','xyz@example.com'],
)
msg.make_private = True
django-sendgrid-v5
into your Django application on his site: https://simpleit.rocks/python/django/adding-email-to-django-the-easiest-way/django-sendgrid-v5
to make a contact form for your web application: https://rx-36.life/create-a-contact-form-using-sendgrid-with-django/