Django - send email on model change - django

I want to send an email when a specific field is changed in a model. Is it possible? Here is what I am looking for. I have a profile model that includes a BooleanField that when the administrator selects to be true I want to send user an email. I know I could put it in a "def save(self):" but, that fires off an email anytime the model is changed and the field is true. Is there a way to have it only email if the field was changed from False to True?

save method is a perfectly good place for what you want to do:
def save(self):
if self.id:
old_foo = Foo.objects.get(pk=self.id)
if old_foo.YourBooleanField == False and self.YourBooleanField == True:
send_email()
super(Foo, self).save()

You can use django-model-changes to do this without an additional database lookup:
from django.db import models
from django.dispatch import receiver
from django_model_changes import ChangesMixin
class MyModel(ChangesMixin, models.Model):
flag = models.BooleanField()
#receiver(pre_save, sender=MyModel)
def send_email_if_flag_enabled(sender, instance, **kwargs):
if instance.previous_instance().flag == False and instance.flag == True:
# send email

Something like this could help and only sends an email when change from false to true
#models.py
from django.contrib.auth.models import User
from django.db.models import signals
from django.db import models
from django.dispatch import receiver
from django.db.models.signals import pre_save
from django.conf import settings
from django.core.mail import send_mail
#signal used for is_active=False to is_active=True
#receiver(pre_save, sender=User, dispatch_uid='active')
def active(sender, instance, **kwargs):
if instance.is_active and User.objects.filter(pk=instance.pk, is_active=False).exists():
subject = 'Active account'
mesagge = '%s your account is now active' %(instance.username)
from_email = settings.EMAIL_HOST_USER
send_mail(subject, mesagge, from_email, [instance.email], fail_silently=False)

Use hook a function with your models post_save using django signals (http://docs.djangoproject.com/en/dev/ref/signals/#django.db.models.signals.post_save)
In that function use standard django mailing: http://docs.djangoproject.com/en/dev/topics/email/

Related

Adding custom functionality to django router methods

I'm trying to add custom functionality to django router methods.
This is my router that exposes the standard methods on an user.
class UserViewSet(viewsets.ModelViewSet):
queryset = User.objects.all()
serializer_class = UserSerializer
permission_classes = [BasePermission]
I'm validating the user using serializer validation methods.
class UserSerializer(serializers.ModelSerializer):
password = serializers.CharField(write_only=True)
MOBILE_ERROR = 'Mobile number should be 10 digits long and only contain numbers.'
EMAIL_ERROR = 'Incorrect email format'
USERNAME_ERROR = 'Username must be at least 6 characters long and contain only letters and numbers.'
class Meta:
model = User
fields = '__all__'
def validate_mobile(self, value):
regexp = re.compile(r'^[0-9]{10}$')
if regexp.search(value):
return value
raise serializers.ValidationError(self.MOBILE_ERROR)
def validate_email(self, value):
if validate_email(value):
return value
raise serializers.ValidationError(self.EMAIL_ERROR)
def validate_username(self, value):
regexp = re.compile(r'^[a-zA-Z0-9]{6,}$')
if regexp.search(value):
return value
raise serializers.ValidationError(self.USERNAME_ERROR)
And this is my route.
router = DefaultRouter(trailing_slash=False)
router.register(r'user', UserViewSet),
urlpatterns = router.urls
I want to add a method send_activation_code if the user is created successfully. How do I do this?
For such purpose you can use signals. Every time when your app creates new User instance - some action should be performed. In your case you should connect build-in signal post_save and your existed send_activation_code function
Example for your case:
yourapp/signals.py:
from django.contrib.auth.models import User
from django.db.models.signals import post_save
from django.dispatch import receiver
#receiver(post_save, sender=User)
def send_activation_code_signal(sender, instance, created, **kwargs):
if created:
send_activation_code(instance.phone_number)
Also, you need to import signals in your app config file
yourapp/app.py:
from django.apps import AppConfig
from django.utils.translation import ugettext_lazy as _
class YourAppConfig(AppConfig):
name = 'yourproject.yourapp'
verbose_name = _('yourapp')
def ready(self):
import yourproject.yourapp.signals
yourapp/__init__.py:
default_app_config = 'yourproject.yourapp.apps.YourAppConfig'
If you dont need to send code every time User instance created - you can specify more statements, for example:
if created and instance.validated:
send_activation_code(instance.phone_number)
There are some more useful built-in signals in Django, check docs
Django signals docs: https://docs.djangoproject.com/en/3.0/ref/signals/

Trigger an email to admin when new user registers in a Django app

I'd like to trigger a simple plaintext email to admin(s) when a new user registers for my Django app. What's the cleanest way to do this?
You can use a post save signal for this. For example:
# <app>/signals.py:
from django.contrib.auth.models import User
from django.db.models.signals import post_save
from django.dispatch import receiver
from django.core.mail import send_mail
#receiver(post_save, sender=User)
def send_email_to_admin(sender, instance, created, **kwargs):
if created:
send_mail(
'<Subject>User {} has been created'.format(instance.username),
'<Body>A new user has been created',
'from#example.com',
['admin#example.com'],
fail_silently=False,
)
# <app>/apps.py
from django.apps import AppConfig
class YourAppConfig(AppConfig):
name = 'app_name'
verbose_name = _('app_name')
def ready(self):
import .signals # noqa
Here, I am using django's mail sending functionality as example, if you are using that then please make sure you have properly configured the settings.

How to get user in Django signals [duplicate]

I have done the below post_save signal in my project.
from django.db.models.signals import post_save
from django.contrib.auth.models import User
# CORE - SIGNALS
# Core Signals will operate based on post
def after_save_handler_attr_audit_obj(sender, **kwargs):
print User.get_profile()
if hasattr(kwargs['instance'], 'audit_obj'):
if kwargs['created']:
kwargs['instance'].audit_obj.create(operation="INSERT", operation_by=**USER.ID**).save()
else:
kwargs['instance'].audit_obj.create(operation="UPDATE").save()
# Connect the handler with the post save signal - Django 1.2
post_save.connect(after_save_handler_attr_audit_obj, dispatch_uid="core.models.audit.new")
The operation_by column, I want to get the user_id and store it. Any idea how can do that?
Can't be done. The current user is only available via the request, which is not available when using purely model functionality. Access the user in the view somehow.
I was able to do it by inspecting the stack and looking for the view then looking at the local variables for the view to get the request. It feels like a bit of a hack, but it worked.
import inspect, os
#receiver(post_save, sender=MyModel)
def get_user_in_signal(sender, **kwargs):
for entry in reversed(inspect.stack()):
if os.path.dirname(__file__) + '/views.py' == entry[1]:
try:
user = entry[0].f_locals['request'].user
except:
user = None
break
if user:
# do stuff with the user variable
Ignacio is right. Django's model signals are intended to notify other system components about events associated with instances and their respected data, so I guess it's valid that you cannot, say, access request data from a model post_save signal, unless that request data was stored on or associated with the instance.
I guess there are lots of ways to handle it, ranging from worse to better, but I'd say this is a prime example for creating class-based/function-based generic views that will automatically handle this for you.
Have your views that inherit from CreateView, UpdateView or DeleteView additionally inherit from your AuditMixin class if they handle verbs that operate on models that need to be audited. The AuditMixin can then hook into the views that successfully create\update\delete objects and create an entry in the database.
Makes perfect sense, very clean, easily pluggable and gives birth to happy ponies. Flipside? You'll either have to be on the soon-to-be-released Django 1.3 release or you'll have to spend some time fiddlebending the function-based generic views and providing new ones for each auditing operation.
You can do that with the help of middleware. Create get_request.py in your app. Then
from threading import current_thread
from django.utils.deprecation import MiddlewareMixin
_requests = {}
def current_request():
return _requests.get(current_thread().ident, None)
class RequestMiddleware(MiddlewareMixin):
def process_request(self, request):
_requests[current_thread().ident] = request
def process_response(self, request, response):
# when response is ready, request should be flushed
_requests.pop(current_thread().ident, None)
return response
def process_exception(self, request, exception):
# if an exception has happened, request should be flushed too
_requests.pop(current_thread().ident, None)
Then add this middleware to your settings:
MIDDLEWARE = [
....
'<your_app>.get_request.RequestMiddleware',
]
Then add import to your signals:
from django.db.models.signals import post_save
from django.contrib.auth.models import User
from <your_app>.get_request import current_request
# CORE - SIGNALS
# Core Signals will operate based on post
def after_save_handler_attr_audit_obj(sender, **kwargs):
print(Current User, current_request().user)
print User.get_profile()
if hasattr(kwargs['instance'], 'audit_obj'):
if kwargs['created']:
kwargs['instance'].audit_obj.create(operation="INSERT", operation_by=**USER.ID**).save()
else:
kwargs['instance'].audit_obj.create(operation="UPDATE").save()
# Connect the handler with the post save signal - Django 1.2
post_save.connect(after_save_handler_attr_audit_obj, dispatch_uid="core.models.audit.new")
Why not adding a middleware with something like this :
class RequestMiddleware(object):
thread_local = threading.local()
def process_request(self, request):
RequestMiddleware.thread_local.current_user = request.user
and later in your code (specially in a signal in that topic) :
thread_local = RequestMiddleware.thread_local
if hasattr(thread_local, 'current_user'):
user = thread_local.current_user
else:
user = None
For traceability add two attributes to your Model(created_by and updated_by), in "updated_by" save the last user who modified the record. Then in your signal you have the user:
models.py:
class Question(models.Model):
question_text = models.CharField(max_length=200)
pub_date = models.DateTimeField('date published')
created_by = models. (max_length=100)
updated_by = models. (max_length=100)
views.py
p = Question.objects.get(pk=1)
p.question_text = 'some new text'
p.updated_by = request.user
p.save()
signals.py
#receiver(pre_save, sender=Question)
def do_something(sender, instance, **kwargs):
try:
obj = Question.objects.get(pk=instance.pk)
except sender.DoesNotExist:
pass
else:
if not obj.user == instance.user: # Field has changed
# do something
print('change: user, old=%s new=%s' % (obj.user, instance.user))
You could also use django-reversion for this purpose, e.g.
from reversion.signals import post_revision_commit
import reversion
#receiver(post_save)
def post_revision_commit(sender, **kwargs):
if reversion.is_active():
print(reversion.get_user())
Read more on their API https://django-reversion.readthedocs.io/en/stable/api.html#revision-api
You can do a small hack by overriding you model save() method and setting the user on the saved instance as additional parameter. To get the user I used get_current_authenticated_user() from django_currentuser.middleware.ThreadLocalUserMiddleware (see https://pypi.org/project/django-currentuser/).
In your models.py:
from django_currentuser.middleware import get_current_authenticated_user
class YourModel(models.Model):
...
...
def save(self, *args, **kwargs):
# Hack to pass the user to post save signal.
self.current_authenticated_user = get_current_authenticated_user()
super(YourModel, self).save(*args, **kwargs)
In your signals.py:
#receiver(post_save, sender=YourModel)
def your_model_saved(sender, instance, **kwargs):
user = getattr(instance, 'current_authenticated_user', None)
PS: Don't forget to add 'django_currentuser.middleware.ThreadLocalUserMiddleware' to your MIDDLEWARE_CLASSES.
I imagine you would have figured this out, but I had the same problem and I realised that all the instances I create had a reference to the user that creates them (which is what you are looking for)
it's possible i guess.
in models.py
class _M(models.Model):
user = models.ForeignKey(...)
in views.py
def _f(request):
_M.objects.create(user=request.user)
in signals.py
#receiver(post_save, sender=_M)
def _p(sender, instance, created, **kwargs):
user = instance.user
No ?
Request object can be obtained from frame record by inspecting.
import inspect
request = [
frame_record[0].f_locals["request"]
for frame_record in inspect.stack()
if frame_record[3] == "get_response"
][0]
def get_requested_user():
import inspect
for frame_record in inspect.stack():
if frame_record[3] == 'get_response':
request = frame_record[0].f_locals['request']
return request.user
else:
return None
context_processors.py
from django.core.cache import cache
def global_variables(request):
cache.set('user', request.user)
----------------------------------
in you model
from django.db.models.signals import pre_delete
from django.dispatch import receiver
from django.core.cache import cache
from news.models import News
#receiver(pre_delete, sender=News)
def news_delete(sender, instance, **kwargs):
user = cache.get('user')
in settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
'web.context_processors.global_variables',
)

Django signal based on the datetime field value

I'm struggling with the following.
I'm trying to create a custom signal that will trigger when the current time will be equal to the value of my model's notify_on DateTimeField.
Something like this:
class Notification(models.Model):
...
notify_on = models.DateTimeField()
def send_email(*args, **kwargs):
# send email
signals.when_its_time.connect(send_email, sender=User)
After I've read through all docs and I found no information on how to implement such a signal.
Any ideas?
UPDATE:
Less naive approach with ability to discard irrelevant tasks: https://stackoverflow.com/a/55337663/9631956
Ok, thanks to comments by #SergeyPugach I've done the following:
Added a post_save signal that calls a function that adds a task to the celery. apply_async let's you pass eta - estimated time of arrival which can accept DateTimeField directly, that's very convenient.
# models.py
from django.db.models import signals
from django.db import models
from .tasks import send_notification
class Notification(models.Model):
...
notify_on = models.DateTimeField()
def notification_post_save(instance, *args, **kwargs):
send_notification.apply_async((instance,), eta=instance.notify_on)
signals.post_save.connect(notification_post_save, sender=Notification)
And the actual task in the tasks.py
import logging
from user_api.celery import app
from django.core.mail import send_mail
from django.template.loader import render_to_string
#app.task
def send_notification(self, instance):
try:
mail_subject = 'Your notification.'
message = render_to_string('notify.html', {
'title': instance.title,
'content': instance.content
})
send_mail(mail_subject, message, recipient_list=[instance.user.email], from_email=None)
except instance.DoesNotExist:
logging.warning("Notification does not exist anymore")
I will not get into details of setting up celery, there's plenty of information out there.
Now I will try to figure out how to update the task after it's notification instance was updated, but that's a completely different story.
In django's documentation there is two interesting signals that may help you on this task: pre_save and post_save.
It depends on your needs, but let's say you want to check if your model's notify_on is equal to the current date after saving your model (actually after calling the save() or create() method). If it's your case you can do:
from datetime import datetime
from django.contrib.auth.models import User
from django.db import models
from django.dispatch import receiver
from django.db.models.signals import post_save
class Notification(models.Model):
...
# Every notification is related to a user
# It depends on your model, but i guess you're doing something similar
user = models.ForeignKey(User, related_name='notify', on_delete=models.DO_NOTHING)
notify_on = models.DateTimeField()
...
def send_email(self, *args, **kwargs):
"""A model method to send email notification"""
...
#receiver(post_save, sender=User)
def create_notification(sender, instance, created, **kwargs):
# check if the user instance is created
if created:
obj = Notification.objects.create(user=instance, date=datetime.now().date())
if obj.notify_on == datetime.now().date():
obj.send_email()
And you should know, that django signals won't work by they own only if there is an action that triggers them. What this mean is that Django signals won't loop over your model's instances and perform an operation, but django signals will trigger when your application is performing an action on the model connected to a signal.
Bonus: To perform a loop over your instances and process an action regulary you may need an asyncworker with a Queue database (mostly, Celery with Redis or RabbitMQ).

django-registration auto create UserProfile

I'm using django-registration and I'm trying to connect to its signals to automatically create a UserProfile.
Signal definition:
from django.dispatch import Signal
# A new user has registered.
user_registered = Signal(providing_args=["user", "request"])
Signal send by django-registration:
def register(self, request, **kwargs):
"""
Create and immediately log in a new user.
"""
username, email, password = kwargs['username'], kwargs['email'], kwargs['password1']
User.objects.create_user(username, email, password)
# authenticate() always has to be called before login(), and
# will return the user we just created.
new_user = authenticate(username=username, password=password)
login(request, new_user)
signals.user_registered.send(sender=self.__class__,
user=new_user,
request=request)
return new_user
My signal connect:
from registration.signals import *
from core.models import UserProfile
from django.contrib.auth.models import User
def createUserProfile(sender, instance, **kwargs):
UserProfile.objects.get_or_create(user=instance)
user_registered.connect(createUserProfile, sender=User)
Needless to say no UserProfile is being created. What am I missing here?
Thanks a lot!
EDIT: I moved my connect() and its corresponding method to a model.py and still no luck.
New code:
from django.db import models
from django.contrib import auth
from django.contrib.auth import login
from core.forms import AuthForm
from registration.signals import *
from django.contrib.auth.models import User
# Create your models here.
class UserProfile(models.Model) :
user = models.ForeignKey(User, unique=True)
def __unicode__(self):
return self.user.username
def createUserProfile(sender, instance, **kwargs):
print "creating profile"
UserProfile.objects.get_or_create(user=instance)
user_registered.connect(createUserProfile, sender=User)
I'm using Pycharm to debug, and in the very beginning my breakpoint on user_registered.connect() is hit. So I assume that connect() is being registered correctly. However, I still don't see createUserProfile being run. Anything else I'm missing?
Thanks!
ANSWER: Doh. My connect and receiver code was wrong. Correct code:
def createUserProfile(sender, user, request, **kwargs):
UserProfile.objects.get_or_create(user=user)
user_registered.connect(createUserProfile)
Realized it after I read signals.py in django-registration
You need to register (connect) your signal in a module which is imported on server startup. Your file where user_registered.connect(createUserProfile, sender=User)lives is mot likely not imported on startup. From the django docs:
You can put signal handling and
registration code anywhere you like.
However, you'll need to make sure that
the module it's in gets imported early
on so that the signal handling gets
registered before any signals need to
be sent. This makes your app's
models.py a good place to put
registration of signal handlers.
http://docs.djangoproject.com/en/dev/topics/signals/#connecting-receiver-functions
So models.py of your custom app would be a good place (or any other module which is definitely imported on server startup).
Torsten is right: the alternative way is to use decorators as stated in documentation:
from registration.signals import user_registered
# ...
#receiver(user_registered)
def your_function_name_here(sender, user, request, **kwargs):
# your code here
pass
I like this way because it's compact and readable.