CODEWITHSUNDEEP
factory-boy
Chris Curvey
Chris Curvey

Reputation: 10389

is there a way to get FactoryBoy to pass a parameter to the save() method of my Django model?

I have a Django model like this:

class NicePerson(models.Model):
    last_name = models.CharField(max_length=100)

    def save(self, make_irish=False, *args, **kwargs):
        """if make_irish is True, prepend last_name with O'"
        if make_irish:
            self.last_name = "O'" + self.last_name

        super(MyModel, self).save(*args, **kwargs)

And I have a FactoryBoy class to build NicePerson instances

class NicePersonFactory(factory.django.DjangoModelFactory):
    class Meta:
        model = NicePerson

I know I can use these together like so:

nice_person = NicePersonFactory(last_name='Shea')

But how can I pass the "make_irish" parameter to my save() method?

Upvotes: 3

Views: 1976

Answers (1)

Xelnor
Xelnor

Reputation: 3589

factory_boy uses the default manager for creation, i.e it calls NicePerson.objects.create() or NicePerson.objects.get_or_create().

For your example, you could override those managers (in your Model definition):

class NicePersonManager(models.Manager):
    def create(self, *args, **kwargs):
        if kwargs.pop('make_irish', False):
            kwargs.update(...)
        return super().create(*args, **kwargs)

Another option would be to override your factory's _create and _get_or_create methods. For instance:

class NicePersonFactory(factory.django.DjangoModelFactory):
    @classmethod
    def _create(cls, model_class, *args, **kwargs):
        make_irish = kwargs.pop('make_irish', False)
        instance = model_class(**kwargs)
        instance.save(make_irish=make_irish)
        return instance

Upvotes: 6

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