模型妈咪：与单个食谱具有外键关系的多个食谱

Question

I have had this annoyance with ModelMommy for a while but I can't figure out how to do this properly.

Let`s assume a simple relation:

class Organization(models.Model):
    label = models.CharField(unique=True)

class Asset(models.Model):
    organization = models.ForeignKey(Organization)
    label = models.CharField(unique=True)

And recipes:

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')

Now when I make these asset recipes I get an error:

>> 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.

This can be solved by providing asset1's organization as a parameter into asset2's make method:

>> asset1 = asset1_recipe.make()
>> asset2 = asset2_recipe.make(organization=asset1.organization)

But there has to be a simpler, more clean way of doing this.

EDIT

Based on the link in Helgi's answer I've changed all my recipe foreign keys to point to a closure:

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')

And can create as many assets as I like:

>> asset1 = asset1_recipe.make()
>> asset2 = asset2_recipe.make()

Answer 1

This doesn't seem to be possible (at least for now) but this feature has been discussed. See here: Reference to same foreign_key object