Create object when created another object in Django

Question

Is it possible to create object when I created another object in Django?

I have code like this and I would like to create AnotherModel instance firstly and then Model instance (when creating AnotherModel).

class Model(models.Model):
    name = models.CharFiled(max_length=50)

class AnotherModel(models.Model):
    model = models.ForeignKey(Model, on_delete=models.CASCADE)
    description = models.TextField()

I tried use Django signals - pre_save.

@receiver(pre_save, sender=AnotherModel)
def save_model(sender, **kwargs):
    # some code

But I have exception like this:

ValueError: Cannot assign "u'Test": "AnotherModel.resource" must be a "Model" instance.

It looks like you did the # some code the wrong way, since you have set the name of your Model to the model (well resource) of your AnotherModel instead of its primary key. But that being said, Signals are usually not a good idea: lincolnloop.com/blog/django-anti-patterns-signals A lot of ORM calls like .update(..), .bulk_create, etc. can circumvent these. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Jul 11, 2019 at 14:13
@swozny13, can you provide the line of code when you are creating the instance of AnotherModel? — lmiguelvargasf
– lmiguelvargasf, Commented Jul 11, 2019 at 14:29

dani herrera · Accepted Answer · 2019-07-11 14:29:30Z

Is it possible to create object when I created another object in Django?

Yes, you can, do you have two options:

1) Option 1: Overriding save ( Overriding predefined model methods )

from django.db import models

class AnotherModel(models.Model):
    model = models.ForeignKey(Model, on_delete=models.CASCADE)
    description = models.TextField()

    def save(self, *args, **kwargs):
        if self.pk is None:  # create
            self.model = Model.objects.create( name = "some random name" )
        super().save(*args, **kwargs)  # Call the "real" save() method.

2) Option 2: Using pre-save signal:

from django.db.models.signals import pre_save
from django.dispatch import receiver    

@receiver(pre_save, sender=AnotherModel)
def my_handler(sender, **kwargs):
    if sender.pk is None:  # create
        sender.model = Model.objects.create( name = "some random name" )

Important:

The related model must be created before to save current one. In examples I used:

Model.objects.create( name = "some random name" )

But you can create model as a regular way:

m = Model()
m.name = "random name"
m.save()  # <--- important!

lmiguelvargasf · Accepted Answer · 2019-07-11 14:35:03Z

0

Yes, it is possible to create an object when another one is created. I think using a signal is possibly the best way to do it.

from django.db.models.signals import pre_save
from django.dispatch import receiver    

@receiver(pre_save, sender=AnotherModel)
def my_handler(sender, **kwargs):
    if sender.pk is None:
        sender.model = Model.objects.create(name="Name")

you just have to create an instance of AnotherModel so that the implemented signal is raised:

>>> o = AnotherModel.objects.create(description='This is a silly description')

You can check that an instance of Model has been created and assigned to o.model:

>>> o.model

answered Jul 11, 2019 at 14:35

lmiguelvargasf

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