Python 如何模拟类层次结构
我想模拟类的层次结构。例如:Python 如何模拟类层次结构,python,mocking,Python,Mocking,我想模拟类的层次结构。例如: class Base(object): def one(self): return 'one' def two(self): return 'two' class Derived(Base): def three(self): return 'three' 定义基本模拟很简单: from mock import Mock def BaseMock(): mock = Mock()
class Base(object):
def one(self):
return 'one'
def two(self):
return 'two'
class Derived(Base):
def three(self):
return 'three'
定义基本模拟很简单:
from mock import Mock
def BaseMock():
mock = Mock()
mock.one.return_value = 'uno'
mock.two.return_value = 'dos'
return mock
def DerivedMock():
mock = BaseMock()
mock.three.return_value = 'tres'
return mock
上述方法可行,但并不完整。我想使用Mock
参数spec
和name
。我可以像往常一样在BaseMock
中指定它们,但在DerivedMock
中,我必须修改privateMock
属性,这是一种糟糕的形式
是否有合适的方法模拟包含
spec
和name
的类层次结构?要模拟类层次结构,请像往常一样定义基类模拟:
from mock import Mock
def BaseMock():
mock = Mock(spec=Base, name='BaseMock')
mock.one.return_value = 'uno'
mock.two.return_value = 'dos'
return mock
在派生的mock中,实例化基本mock并(尽管它很讨厌)修改与spec
和name
相关的mock私有属性:
def DerivedMock():
mock = BaseMock()
# Set Mock private attributes to Derived
mock._mock_methods.extend(dir(Derived))
mock._mock_name = 'DerivedMock'
mock._spec_class = Derived
mock.three.return_value = 'tres'
# Defining DerivedMock.four would cause a spec error
# mock.four.return_value = 'quatro'
return mock
现在,mock DerivedMock看起来像是类派生的,只是返回值不同:
d = DerivedMock()
assert isinstance(d, Derived)
assert repr(d).startswith("<Mock name='DerivedeMock' spec='Derived'")
assert d.one() == 'uno'
assert d.two() == 'dos'
assert d.three() == 'tres'
d=DerivedMock()
断言isinstance(d,派生)
断言repr(d).startswith("为什么你提出的方法不好?在每种方法中,你都需要构建一个Mock
,修改one
、two
和two
。那么为什么你当前的方法不好呢?@Simeon Visser:上面给出的方法是不完整的,因为DerivedMock
没有使用Mockspecde>和名称
参数。