I'm using factory_boy to replace fixtures in a Django app. I have a Product model that should have many Offers and Merchants.
#models.py
class Product(models.Model):
name = models.CharField()
class Merchant(models.Model):
product = models.ForeignKey(Product)
name = models.CharField()
class Offer(models.Model):
product = models.ForeignKey(Product)
price = models.DecimalField(max_digits=10, decimal_places=2)
I want a factory that creates a Product with several Merchants and several Offers.
#factories.py
import random
from models import Offer, Merchant, Product
class OfferFactory(factory.django.DjangoModelFactory):
FACTORY_FOR = Offer
product = factory.SubFactory(ProductFactory)
price = random.randrange(0, 50000, 1)/100.0
class MerchantFactory(factory.django.DjangoModelFactory):
FACTORY_FOR = Merchant
product = factory.SubFactory(ProductFactory)
name = factory.Sequence(lambda n: 'Merchant %s' % n)
url = factory.sequence(lambda n: 'www.merchant{n}.com'.format(n=n))
class ProductFactory(factory.django.DjangoModelFactory):
FACTORY_FOR = Product
name = "test product"
offer = factory.RelatedFactory(OfferFactory, 'product')
offer = factory.RelatedFactory(OfferFactory, 'product') # add a second offer
offer = factory.RelatedFactory(OfferFactory, 'product') # add a third offer
merchant = factory.RelatedFactory(MerchantFactory, 'product')
merchant = factory.RelatedFactory(MerchantFactory, 'product') # add a second merchant
merchant = factory.RelatedFactory(MerchantFactory, 'product') # add a third merchant
But when I use ProductFactory to create a Product, it only has one offer and one merchant.
In [1]: from myapp.products.factories import ProductFactory
In [2]: p = ProductFactory()
In [3]: p.offer_set.all()
Out[3]: [<Offer: $39.11>]
How do I set up a ProductFactory to have more than one dependent of a particular type?
To be able to specify the number of related objects in parent factory:
class Company(models.Model):
name = models.CharField(max_length=255)
class ContactPerson(models.Model):
name = models.CharField(max_length=255)
company = models.ForeignKey(Company, on_delete=CASCADE, related_name='contacts')
class CompanyFactory(factory.django.DjangoModelFactory):
name = factory.Faker('company')
class Meta:
model = Company
@factory.post_generation
def add_contacts(self, create, how_many, **kwargs):
# this method will be called twice, first time how_many will take the value passed
# in factory call (e.g., add_contacts=3), second time it will be None
# (see factory.declarations.PostGeneration#call to understand how how_many is populated)
# ContactPersonFactory is therefore called +1 times but somehow we get right amount of objs
at_least = 1
if not create:
return
for n in range(how_many or at_least):
ContactPersonFactory(contact=self)
class ContactPersonFactory(factory.django.DjangoModelFactory):
name = factory.Faker('first_name')
class Meta:
model = ContactPerson
company = CompanyFactory(company_name='ACME ltd', add_contacts=4)
print(repr(company.name), len(company.contacts.all()))
company = CompanyFactory(company_name='ACME ltd')
print(repr(company.name), len(company.contacts.all()))
---
'ACME ltd' 4
'ACME ltd' 1
If you are ok with always just one child, the docs solution works well:
class CompanyFactory(factory.django.DjangoModelFactory):
name = factory.Faker('company')
whatever_really = factory.RelatedFactory('my_app.factories.ContactPersonFactory', 'contact')
class Meta:
model = Company
note the full path to the related factory.
company = CompanyFactory(company_name='ACME ltd')
print(repr(company.name), len(company.contacts.all()))
---
'ACME ltd' 1
$ pip freeze | egrep 'factory|Faker|Django'
Django==2.0.4
factory-boy==2.10.0
Faker==0.8.13
$ python -V
Python 3.6.5