模型妈咪:与单个食谱具有外键关系的多个食谱

时间:2016-08-23 10:21:55

标签: django testing model-mommy

我对ModelMommy有一段时间的烦恼,但我无法弄清楚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,我收到一个错误:

>> 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asset2的make方法的参数来解决:

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

但必须有一种更简单,更干净的方法。

修改

根据Helgi的回答中的链接,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')

可以创建尽可能多的资产:

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

1 个答案:

答案 0 :(得分:0)

这似乎不可能(至少目前为止),但已讨论过此功能。见这里:Reference to same foreign_key object