Question

我认为不是以ATTR__SUBATTR的形式将数据传递给工厂的RelatedFactory，您也应该能够直接传递已经存在的实例。除非我遗漏了一些非常明显的东西，否则这似乎不起作用。看看：

class Owner(models.Model):
    name = models.CharField()

class Item(models.Model):
    name = models.CharField()
    owner = models.ForeignKey(Owner, null = True, related_name = 'items')

class ItemFactory(factory.django.DjangoModelFactory):
    class Meta:
        model = Item

class OwnerFactory(factory.django.DjangoModelFactory):
    class Meta:
        model = Owner
    items = factory.RelatedFactory(ItemFactory, 'owner')

item = Item.objects.create(name='Foo')
alice = OwnerFactory(name='Alice', items__name='Bar')
alice.items.all()
<QuerySet [<Item: Bar>]>
bob = OwnerFactory(name='Bob', items=item) # or items = [item] doesn't matter
bob.items.all()
<QuerySet []>

一直在努力让我的工厂变得更好，干涸并且遇到了这个障碍。写了我自己的RelatedFactory改编版，它允许一次处理多个值，如果你在这个过程中创建新对象，它可以正常工作 - 但是如果你使用已经存在的对象则不行。

有效的示例：OwnerFactory(items__name=['Foo','Bar'])=> Foo and Bar in owner.items。
不起作用的示例：OwnerFactory(items=[foo,bar])=> owner.items is empty
请注意，我在顶部的大例子中使用了默认的RelatedFactory。

我整天都在查看factory_boy的文档，但是现在找不到解决方案和隧道视图，禁止任何新见解。

Answer 1

您正在寻找http://factoryboy.readthedocs.io/en/latest/recipes.html#simple-many-to-many-relationship

class ItemFactory(factory.django.DjangoModelFactory):
    class Meta:
        model = Item

class OwnerFactory(factory.django.DjangoModelFactory):
    class Meta:
        model = Owner

    @factory.post_generation
    def items(self, create, extracted, **kwargs):
        if not create:
            # Simple build, do nothing.
            return

        if extracted:
            # A list of items were passed in, use them
            for item in extracted:
                self.items.add(item)

factory_boy：传递RelatedFactory的模型实例

1 个答案: