Question

我正在为一个模型工厂工作，我正在尝试填充一个有选择列表的字段。当我尝试使用Factory创建一个对象时，我尝试从选项列表中填写一个随机选项，抛出异常：

TypeError：＆＃39; choice＆＃39;是此函数的无效关键字参数

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/usr/local/lib/python2.7/dist-packages/factory/base.py", line 551, in build
    return cls._generate(enums.BUILD_STRATEGY, kwargs)
  File "/usr/local/lib/python2.7/dist-packages/factory/base.py", line 505, in _generate
    return step.build()
  File "/usr/local/lib/python2.7/dist-packages/factory/builder.py", line 279, in build
    kwargs=kwargs,
  File "/usr/local/lib/python2.7/dist-packages/factory/base.py", line 312, in instantiate
    return self.factory._build(model, *args, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/factory/base.py", line 531, in _build
    return model_class(*args, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/django/db/models/base.py", line 571, in __init__
    raise TypeError("'%s' is an invalid keyword argument for this function" % list(kwargs)[0])
TypeError: 'choice' is an invalid keyword argument for this function

使用的版本：

django==1.11
factory-boy==2.9.2
python==2.7.12

（裁剪）模型：

class Server(models.Model):

    TEST = 'test'
    ACCEPT = 'accept'

    SERVER_TYPES = (
        (TEST, _("Testing Server")),
        (ACCEPT, _("Acceptation Server"))
    )

    type = models.CharField(_("Server type"), max_length=50, choices=SERVER_TYPES)

（裁剪）工厂：

class ServerFactory(factory.DjangoModelFactory):

    type = factory.Faker('random_element', elements=[choice[0] for choice in Server.SERVER_TYPES)

    class Meta:
        model = Server

我没有使用Faker('random_element, elements=[..])，而是尝试使用LazyFunction：

def get_server_type():
    choices = [choice[0] for choice in Server.SERVER_TYPES]
    return random.choice(choices)

class ServerFactory(factory.DjangoModelFactory):

    organization = factory.SubFactory(OrganizationFactory)
    type = factory.LazyFunction(get_server_type)

    .. Meta ..

这也会引发同样的错误。我也找不到任何其他替代方法来解决这个问题。有关使用type包时如何使用SERVER_TYPES个选项填充factory字段的任何建议吗？

Answer 1

你能尝试一下吗？

from random import choice
type = factory.LazyAttribute(lambda x: choice(Server.SERVER_TYPES)[0])

基于对问题的初步描述的旧评论：

应为type = factory.Faker('random_element', elements=[choice[0] for choice in Server.SERVER_TYPES])

Django FactoryBoy：使用选项填充modelfield会引发错误

1 个答案: