我有一段时间对 ModelMommy 感到烦恼，但我不知道如何正确地做到这一点。
让我们假设一个简单的关系：
class Organization(models.Model):
label = models.CharField(unique=True)
class Asset(models.Model):
organization = models.ForeignKey(Organization)
label = models.CharField(unique=True)
和食谱：
from model_mommy.recipe import Recipe, foreign_key
organization_recipe = Recipe(Organization, label='My Organization')
asset1_recipe = Recipe(Asset,
organization=foreign_key(organization_recipe),
label='asset 1')
asset2_recipe = Recipe(Asset,
organization=foreign_key(organization_recipe),
label='asset 2')
现在，当我制作这些资产配方时，我得到一个错误：
>> asset1 = asset1_recipe.make()
>> asset2 = asset2_recipe.make()
IntegrityError: duplicate key value violates unique constraint "organizations_organization_label_key"
DETAIL: Key (label)=(My Organization) already exists.
这可以通过将asset1的组织作为参数提供给asset2的make方法来解决：
>> asset1 = asset1_recipe.make()
>> asset2 = asset2_recipe.make(organization=asset1.organization)
但是必须有一种更简单、更干净的方式来做到这一点。
编辑
根据 Helgi 回答中的链接，我已将所有配方外键更改为指向闭包：
def organization_get_or_create(**kwargs):
"""
Returns a closure with details of the organization to be fetched from db or
created. Must be a closure to ensure it's executed within a test case
Parameters
----------
kwargs
the details of the desired organization
Returns
-------
Closure
which returns the desired organization
"""
def get_org():
org, new = models.Organization.objects.get_or_create(**kwargs)
return org
return get_org
my_org = organization_get_or_create(label='My Organization')
asset1_recipe = Recipe(Asset,
organization=my_org,
label='asset 1')
asset2_recipe = Recipe(Asset,
organization=my_org,
label='asset 2')
并且可以创建任意数量的资产：
>> asset1 = asset1_recipe.make()
>> asset2 = asset2_recipe.make()