CODEWITHSUNDEEP
djangofactory-boy
sebap123
sebap123

Reputation: 2685

How to resolve CyclicDefinitionError in factory_boy SubFactory call?

I have following models

# in ModelA_App/models.py
class ModelA(models.Model):
    TYPEA = 1
    TYPEB = 2
    TYPE_CHOICES = (
       (TYPEA, 'TypeA'),
       (TYPEB, 'TypeB')
    )
    type = models.PositiveSmallIntegerField(choices=TYPE_CHOICES)
    name - models.CharField(max_length = 100)

#in ModelB_App/models.py
from ModelA_App.models import ModelA

class ModelB(models.Model):
    label = models.TextFiled()
    model_a = models.OneToOneField(ModelA, on_delete=models.CASCADE)

And I have following factories:

#in ModelA_App/factories.py
class ModelAFactory(factory.django.DjangoModelFactory):
    class Meta:
       model = ModelA

    name = factory.Faker('word')
    type = ModelA.TYPEA

#in ModelB_App/factories.py
from ModelA_App.models import ModelA
from ModelA_App.factories import ModelAFactory

class ModelBFactory(factory.django.DjangoModelFactory):
    class Meta:
       model = ModelB

    label = factory.Faker('text')
    model_a = SubFactory(ModelAFactory, type = factory.LazyAttribute(lambda o: '%d' % o.type))

    class Params:
       type = ModelA.TYPEA

I'd like to be able to create ModelB object with ModelA having TYPEB. Trying line ModelBFactory.create(type = ModelA.TYPEB) results in error:

factory.errors.CyclicDefinitionError: Cyclic lazy attribute definition for 'type'; cycle found in ['type']

Additionally, when I change type name in Params class to e.g. model_type having:

model_a = SubFactory(ModelAFactory, type = factory.LazyAttribute(lambda o: '%d' % o.model_type))

It fails with

AttributeError: The parameter 'model_type' is unknown.

How can I achieve my goal?

Upvotes: 3

Views: 1051

Answers (2)

Xelnor
Xelnor

Reputation: 3609

When you write a SubFactory, the additional definitions you provide are anchored to that sub-factory. What you have written is equivalent to:

class SubModelAFactory(ModelAFactory):
    class Meta:
        # Not required - implied by class inheritance
        model = models.ModelA

    type = factory.LazyAttribute(lambda o: '%d' % o.type)

class ModelBFactory(factory.django.DjangoModelFactory):
    ...
    model_a = factory.SubFactory(SubModelFactory)

Instead, you should go "up" a level, as described in the SelfAttribute docs:

class ModelBFactory(factory.django.DjangoModelFactory):
    ...
    model_a = factory.SubFactory(
        SubModelFactory,
        type=factory.LazyAttribute(lambda o: '%d' % o.factory_parent.type),

        # If the value can be passed without conversion, use:
        type=factory.SelfAttribute('..type'),
    )

Upvotes: 3

sebap123
sebap123

Reputation: 2685

I've found solution. ModelBFactory should look like this:

#in ModelB_App/factories.py
from ModelA_App.models import ModelA
from ModelA_App.factories import ModelAFactory

class ModelBFactory(factory.django.DjangoModelFactory):
    class Meta:
       model = ModelB

    label = factory.Faker('text')
    model_a = factory.LazyAttribute(lambda o: ModelAFactory(type = o.type))

    class Params:
       type = ModelA.TYPEA

Instead of putting ModelAFactory attribute as a LazyAttribute I had to put whoel factory as one.

Upvotes: 0

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