我试图使用factory_boy和faker用一些随机数据填充我的Django项目。但是,当我尝试创建UserFactory对象的实例时,我得到错误TypeError :()缺少1个必需的位置参数:' a'。
它似乎与fake_date和date_joined属性及其lambda函数有关。我发现的一些教程已经以这种方式为Factory对象使用lambda函数进行了布局,但它并不适用于我。
import datetime
import factory
import faker
from dataStoreApp.models import Category, Goal, Transaction
from django.contrib.auth.models import User
fake = faker.Faker()
class CategoryFactory(factory.django.DjangoModelFactory):
class Meta:
model = Category
name = factory.Iterator(['Category1', 'Category2',
'Category3', 'Category4',
'Category5', ], cycle=False)
class UserFactory(factory.django.DjangoModelFactory):
FACTORY_HIDDEN_ARGS = ('fake_date', )
class Meta:
model = User
fake_date = factory.LazyAttribute(
lambda a: datetime.datetime.now().strftime("%Y-%m-%d %H:%M"))
date_joined = factory.LazyFunction(lambda a: a.fake_date)
first_name = factory.LazyAttribute(lambda b: fake.first_name())
last_name = factory.LazyAttribute(lambda c: fake.last_name())
email = factory.LazyAttribute(lambda e: 'test' + '@test_email.com')
class GoalFactory(factory.DjangoModelFactory):
class Meta:
model = Goal
class TransactionFactory(factory.django.DjangoModelFactory):
class Meta:
model = Transaction
在python控制台中运行时,我收到以下错误:
Traceback (most recent call last):
File "<input>", line 1, in <module>
File "C:\Program Files\Python36\lib\site-packages\factory\base.py", line 568, in create
return cls._generate(enums.CREATE_STRATEGY, kwargs)
File "C:\Program Files\Python36\lib\site-packages\factory\base.py", line 505, in _generate
return step.build()
File "C:\Program Files\Python36\lib\site-packages\factory\builder.py", line 272, in build
step.resolve(pre)
File "C:\Program Files\Python36\lib\site-packages\factory\builder.py", line 221, in resolve
self.attributes[field_name] = getattr(self.stub, field_name)
File "C:\Program Files\Python36\lib\site-packages\factory\builder.py", line 363, in __getattr__
extra=declaration.context,
File "C:\Program Files\Python36\lib\site-packages\factory\declarations.py", line 59, in evaluate
return self.function()
TypeError: <lambda>() missing 1 required positional argument: 'a'
答案 0 :(得分:6)
您可以在代码
中创建带参数a
的函数
lambda a: ....
但似乎程序期望没有参数的函数
lambda: ...