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Home/ Questions/Q 9172357
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T16:19:36+00:00 2026-06-17T16:19:36+00:00

I have the following code: class Album(models.Model): name = models.CharField(max_length=255, unique=True, null=False) rating =

  • 0

I have the following code:

class Album(models.Model):
  name = models.CharField(max_length=255, unique=True, null=False)
  rating = models.ForeignKey("Rating", null=False)

class Rating(models.Model):
  value = models.IntegerField(null=False, default=0)

What is the best way (in the django/python philosophy) to create an object (Album) and it’s sub object (Rating) and save it?

I have done this:

a = Album()
a.name = "..."
r = Rating()
r.save()
a.rating = r
a.save()

I don’t like this because the part of creating the sub object empty is totally not useful.
I’d prefer some simple way like this – the sub-object should be created automatically:

a = Album()
a.name = "..."
a.save()
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T16:19:37+00:00Added an answer on June 17, 2026 at 4:19 pm

    You’ll want to look into signals.

    Essentially a signals are sent when an Object is saved.

    Using a pre_save signal you can then create a Rating and associate it to the new Album jsut before it is saved for the first time.

    from django.db.models.signals import pre_save
    from django.dispatch import receiver
    from myapp.models import Album, Rating
    
    @receiver(pre_save, sender=Album)
    def add_rating_to_album(sender, **kwargs):
    
        # If saving a new Album
        if not instance.id:
    
            # Create and save a new rating
            rating = Rating()
            rating.save()
    
            # Associate it to the Album being saved
            instance.rating = rating
    
        # Continue to normal save with new rating applied
    

    I haven’t tested this specific code but it should get you in the right direction

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