使用 Python 模拟模块，我如何修补 class 以便它只存根我想存根的方法而单独保留其他属性和方法？

Question

我想要做的是在 class 中存根几个方法，这些方法在我正在测试的某些代码中间接实例化。我希望那个补丁 class 的所有其他属性和方法都能正常运行。

这是一个简单的案例，它展示了我想要的东西 (Python 2.7)。（注意：在我的实际用例中，MyClass 是一个 class，它在我正在测试的某些代码中间接实例化）：

from mock import patch

class MyClass(object):
    def __init__(self):
        self.prop = 'prop'

    def foo(self):
        return 'foo'

    def bar(self):
        return 'bar'

    def unmocked(self):
        return 'unmocked'

patcher = patch('__main__.MyClass')
MockedClass = patcher.start()
instance = MockedClass.return_value
instance.foo.return_value = 'mocked foo!'
instance.bar.return_value = 'mocked bar!'

my_instance = MyClass()
assert my_instance.foo() == 'mocked foo!', my_instance.foo()
assert my_instance.bar() == 'mocked bar!', my_instance.bar()

# These asserts fail
assert my_instance.unmocked() == 'unmocked', my_instance.unmocked()
assert my_instance.prop == 'prop', my_instance.prop

patcher.stop()

结果：

$ python mock_test.py
Traceback (most recent call last):
  File "mock_test.py", line 27, in <module>
    assert my_instance.unmocked() == 'unmocked', my_instance.unmocked()
AssertionError: <MagicMock name='MyClass().unmocked()' id='140400215338448'>

为什么 unmocked 方法现在返回一个 MagicMock 对象？我怎样才能修补这个 class 以便它只存根我想要的方法而不管其他所有东西？

Answer 1

您应该使用 patch.multiple:

修补您的 class

import mock

patcher = mock.patch.multiple(
    '__main__.MyClass',
    foo=mock.Mock(return_value='mocked foo!'),
    bar=mock.Mock(return_value='bar')
)

用上面的代码更新代码后，它将运行并且所有断言都通过。这是一个完整的工作示例：

from mock import patch, Mock

class MyClass(object):
    def __init__(self):
        self.prop = 'prop'

    def foo(self):
        return 'foo'

    def bar(self):
        return 'bar'

    def unmocked(self):
        return 'unmocked'

patcher = patch.multiple('__main__.MyClass',
                         foo=Mock(return_value='mocked foo!'),
                         bar=Mock(return_value='mocked bar!'))
patcher.start()

my_instance = MyClass()
assert my_instance.foo() == 'mocked foo!', my_instance.foo()
assert my_instance.bar() == 'mocked bar!', my_instance.bar()
assert my_instance.unmocked() == 'unmocked', my_instance.unmocked()
assert my_instance.prop == 'prop', my_instance.prop

patcher.stop()

Answer 2

from mock import patch

class MyClass(object):
    def __init__(self):
        self.prop = 'prop'

    def foo(self):
        return 'foo'

    def bar(self):
        return 'bar'

patcher = patch.object(MyClass,"foo",return_value='mocked foo!')
MockedClass = patcher.start()


my_instance = MyClass()
assert my_instance.foo() == 'mocked foo!', my_instance.foo()

# These asserts will fail
assert my_instance.bar() == 'bar', my_instance.bar()
assert my_instance.prop == 'prop', my_instance.prop

patcher.stop()

您可以使用 patch.object

直接模拟您想要的 method 而不是整个 class

使用 Python 模拟模块，我如何修补 class 以便它只存根我想存根的方法而单独保留其他属性和方法？

Using the Python mock module, how can I patch a class so that it stubs only the methods I want to stub and leaves other properties and methods alone?

python

unit-testing

mocking