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
