Django: use related_name to create a subclass

Question

How can I be able to use related_name to create a child model when the parent model is being created.

from django.db import models

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


class Parent(models.Model):
    name = models.TextField()
    child = models.ForeignKey(Child, on_delete=models.SET_NULL, related_name="parent_child", null=True, blank=True)

    def save(self, *args, **kwargs):
        super().save()

@receiver(post_save, sender=Parent)
def post_save_receiver(sender, instance, *args, **kwargs)
    if not instance.child:
        instance.child.parent_child.create(name="I am a child")


class Child(models.Model):
    name = models.CharField()

Currently, on save, I get a NoneType error:

'NoneType' object has no attribute 'parent_child'

These are the local variables in my stack trace

args: ()

instance: <Parent: asdf - asd-47013855>

kwargs:
{
 'created': True,
 'raw': False,
 'signal': <django.db.models.signals.ModelSignal object at 0x7f20b92e2ce0>,
 'update_fields': None,
 'using': 'default'
}

sender: <class 'parent.models.Parent'>

You're trying to create an object inside `save()` method of `Parent` which means you're trying to create child before parent is created — Ankit Tiwari, Aug 03 '22 at 06:33
Makes sense, what is the correct / recommended way to go about it? Can I use a `post_save` receiver? — Newton Karanu, Aug 03 '22 at 06:35
You can use [signals](https://docs.djangoproject.com/en/4.0/ref/signals/#signals) — Ankit Tiwari, Aug 03 '22 at 06:37
@AnkitTiwari updated my question to include `signals`. I still get the same error — Newton Karanu, Aug 03 '22 at 07:03

Ankit Tiwari · Accepted Answer · 2022-08-03T16:06:14.473

1

You're trying to perform action on NoneType object eg. instance.child because when you create your Parent instance you've to assing child to your Parent.child attribute so by default it will None so to assign child to your Parent class you can do like this & you don't require signals you can do this inside your save method.

class Parent(models.Model):
    name = models.TextField()
    child = models.ForeignKey(Child, on_delete=models.SET_NULL, related_name="parent_child", null=True, blank=True)

    def save(self, *args, **kwargs):
        if not self.child:
           self.child = Child.objects.create(name="I am a child")
        super(Parent, self).save(*args, **kwargs)

also you're trying to access related object in wrong way if you're inside Parent you can access child directly using Parent.child & if you want to access Parent from Child then you can do like this Child.parent_child read more about backward relation

edited Aug 03 '22 at 16:06

answered Aug 03 '22 at 07:17

Ankit Tiwari

4,438
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you miss if self.child is None: , except that evrything is OK – Waldemar Podsiadło Aug 03 '22 at 07:19
He is trying to create child when Parent is created so by default child will be None – Ankit Tiwari Aug 03 '22 at 07:21
1

yes, but your save overrides child with new evrytime Parent is saved(even when only edited) – Waldemar Podsiadło Aug 03 '22 at 07:26

score 0 · Answer 2 · answered Aug 03 '22 at 07:17

0

def post_save_receiver(sender, instance, *args, **kwargs)
    if not instance.child:
        child = Child(name='I am a child').save()
        instance.child = child
        instance.save()

answered Aug 03 '22 at 07:17

Waldemar Podsiadło

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Django: use related_name to create a subclass

2 Answers2