[FIXED] Django model referring to self throws errors

Issue

I have a problem to define Django model where the Primary Key is UUID5 that base on model fields.

Here is what I have got already.

class Place(models.Model):
    uuid = models.UUIDField(primary_key=True, default=uuid.uuid5('PLACE_UUID_NAMESPACE', '{}{}{}'.format(self.city, self.zip, self.country)))
    city = models.CharField(max_length=24, blank=True, null=True)
    zip = models.CharField(max_length=6, blank=True, null=True)
    country = CountryField(default='US')

The problem is with referring to model field as I did self.city, self.zip, self.country. Django does not accept the way I did. How can I define the uuid5 using model fields?

I appreciate your help.

Solution

The problem is with referring to model field as I did self.city, self.zip, self.country. Django does not accept the way I did.

Yes and Django is right. You do not have an access to self yet until you have an instance of the class. So try this instead:

class Place(models.Model):
    uuid = models.UUIDField(primary_key=True, default=uuid.uuid4)
    •••

So that if we instantiate an instance with for example:

aPlace = Place(city=“Bournemouth”, zip=“blah”)

Now we have aPlace Which has by default, a random but unique identifier (UUID) in it’s uuid field and a CountryField(default='US') in it’s country field. Also city field is “Bournemouth”, and zip is “blah”.

Then you can manipulate the instance right before saving to database. That is where you change certain fields for example the uuid field.

aPlace.uuid = uuid.uuid5('PLACE_UUID_NAMESPACE', '{}{}{}'.format(aPlace.city, aPlace.zip, aPlace.country))
aPlace.save()

Or directly override the save method of the model:

def save(self, *args, **kwargs):
    self.uuid = uuid.uuid5('PLACE_UUID_NAMESPACE', '{}{}{}'.format(self.city, self.zip, self.country))
    super().save(*args, **kwargs)

Answered By - Chukwujiobi Canon