I want to call a method after this object (in this example Question) is saved.
This is what I got:
class Question(models.Model):
...
def after_save(sender, instance, *args, **kwargs):
some_method(instance)
post_save.connect(Question.after_save, sender=Question)
It kind of works. But the problem is that the instance data is the old one (same as before the save). So some_method(instance) gets the old instance.
But I need the instance with the new data.
The method you call from post_save signal should be outside of the Model. You can put the method inside the models.py or in another file such as signals.py
class Question(models.Model):
...
def some_method(self):
return "hello"
def question_saved(sender, instance, *args, **kwargs):
instance.some_method()
post_save.connect(question_saved, sender=Question)
You can override save method, and call what ever you want after the object gets saved
Here is a related link: Django. Override save for model
What you want is something called signals: https://docs.djangoproject.com/en/3.2/topics/signals/.
Some operations send a signals before and after they're completed successfully. In the docs you can find a list of built-in signals. You can also create custom signals. For example assume that I want to do something specific each time a new user is created. Then my code would be something like this:
users/models.py
from django.contrib.auth.models import AbstractUser
from django.dispatch import receiver
from django.db.models.signals import post_save
class CustomUser(AbstractUser):
pass
###############
# Recieve the signals and get the signals that are sent
# after models are saved and whose sender is CustomUser
###############
#receiver(post_save, sender='CustomUser')
def do_something():
do_something_specific
You can also save your signals in a separate file called signals.py but you need to change your app's apps:
<appname>/apps.py
from django.apps import AppConfig
class AppnameConfig(AppConfig):
name = <appname>
def ready(self):
import <appname>.signals
Related
I am new to python and django and I want to know the best way to set things up.
I have a model called OauthProviders which I set up to encrypt some fields before save in the ModelViewSet (override perform_create method). I dont want to create routes (urls) for this model.
Now, if I want to access this model in the code (I can with OauthProvider.objects.all() of course), but I have a few questions:
how do I enter data to this model NOT in code? If I use the admin portal for it, it doesn't execute my custom perform_create method, so it gets added to the database in plain text
What is the best way to decrypt a message if I retrieve data?
EDIT:
I moved the logic from the ModelViewSet to the save() method on the model with the following code:
def save(self, *args, **kwargs):
self.credentials = encrypt_message(self.credentials, '<keyhere>')
return super().save(*args, **kwargs)
This seems to work. Is this good practice? How to do this with the get method?
A good way of doing this is using django signals.
Assuming you have:
class OauthProvider(models.Model):
pass # some fields here
In your case you can use either:
pre_save - code that will be executed before OauthProvider gets saved in the database.
post_save - code that will be executed after OauthProvider gets saved in the database.
It's up to you to decide when exactly you want to execute additional code.
Example with pre_save:
from django.db.models.signals import pre_save
from django.dispatch import receiver
class OauthProvider(models.Model):
pass # some fields here
#receiver(pre_save, sender=OauthProvider)
def oauthprovider_on_save(sender, instance, **kwargs):
instance.credentials = encrypt_message(self.credentials, '<keyhere>')
# no need to call instance.save() in pre_save signal
The pre_save signal will be executed:
If you call obj.save() programatically.
If you save the obj through django admin site.
If you create/update the obj within rest api
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).
Sitting over a day on it. Really can't understand why this signal is not triggered when a user is activated, no error log, no exception in the admin on activation. Can anybody help? The following code should result in a log message in the apache error.log when a user, right?
import logging
from django.dispatch import receiver
from registration.signals import user_activated
#receiver(user_activated)
def registered_callback(sender, **kwargs):
logger = logging.getLogger("user-activated")
logger.error("activated here")
same with user_registered
First of all im using django 1.8.3 .You should register your signal first. As far as i know, there are some methods to do that but this is what im doing;
Create signals.py in your app write your signal there;
from django.db.models.signals import post_save
from django.dispatch import receiver
#receiver(post_save, sender=your_model,dispatch_uid="yourmodel_save_receiver")
def post_save_yourmodel(sender, instance, **kwargs):
if instance.profile_status:
print "active"
else:
print "not active"
Then you should create apps.py. This file contains configuration information to your model.
from django.apps import AppConfig
class yourmodel_config(AppConfig):
name = 'yourmodel_config'
verbose_name = 'your_model config'
def ready(self):
import yourmodel.signals
With this whenever your app is ready, your signals will be imported
Finally open your __init__.py and add the following.
default_app_config = 'yourmodel.apps.yourmodel_config'
With this you are defining application configuration for your model.This example when ever yourmodel is saved, signal checks for profile_status attribute and prints output depending on the value(true or false) to your console. You can also add created parameter to your model to know that if instance of the model is created. created will return True if a new record was created. def post_save_yourmodel(sender, instance, created, **kwargs):. Otherwise this signal will be triggered whenever your model is saved with yourmodel.save().
Consider that is a post_save example.You can find list of the model signals from here.
I want to execute some code in a Django model when it is first created. After that whenever it is saved I want to execute some other code. The second task can be easily done by overriding the save() method.
How can I do the first task?
Extending sdolan's answer by using receiver decorator:
from django.db import models
from django.dispatch import receiver
class MyModel(models.Model):
pass
#receiver(models.signals.post_save, sender=MyModel)
def execute_after_save(sender, instance, created, *args, **kwargs):
if created:
# code
You can use django signals' post_save:
# models.py
from django.db.models import signals
class MyModel(models.Model):
pass
def my_model_post_save(sender, instance, created, *args, **kwargs):
"""Argument explanation:
sender - The model class. (MyModel)
instance - The actual instance being saved.
created - Boolean; True if a new record was created.
*args, **kwargs - Capture the unneeded `raw` and `using`(1.3) arguments.
"""
if created:
# your code goes here
# django 1.3+
from django.dispatch import dispatcher
dispatcher.connect(my_model_post_save, signal=signals.post_save, sender=MyModel)
# django <1.3
from django.db.models.signals import post_save
post_save.connect(my_model_post_save, sender=MyModel)
field.default
The default value for the field. This can be a value or a callable object. If callable it will be called every time a new object is created.
So, we can use any field callable to do what we want on creation, huh? ;-)
I'm trying to override the default User model in Django to add some logic into the save() method. I'm having a hard time trying to figure out out to go about this.
I'm using Django 1.1 if that helps.
I used post_save since i need to add the user into ldap.. I just added this into a models.py
from django.db import models
from django.contrib.auth.models import User
from django.db.models import signals
from django.dispatch import dispatcher
def user_post_save(sender, instance, **kwargs):
print "got here"
models.signals.post_save.connect(user_post_save, sender=User)
Don't. Instead catch the pre_save signal.
You'd better use a Proxy model, so to use the same table but overriding behavior.
This is the standard way to extend Django's own models, because they cannot be made abstract.
So declare your model as:
from django.contrib.auth.models import User
class CustomUser(User):
class Meta:
proxy = True
def save(self, *args, **kwargs):
# do anything you need before saving
super(CustomUser, self).save(*args, **kwargs)
# do anything you need after saving
and you are done.