难以理解如何模拟一个类,并能够断言它的'方法是用一些参数调用的。当我断言调用时,我得到一个“未调用”断言但是,我可以在mock_calls
属性中看到方法调用。
sandbox/module.py
class Subject(object):
def __init__(self):
pass
def run(self, *args, **kwargs):
reference = Reference(*args, **kwargs)
reference.method_a(*args)
class Reference(object):
def __init__(self, *args, **kwargs):
pass
def method_a(self, *args):
pass
test.py
import unittest
from unittest import mock
from sandbox.module import Subject
class TestSandbox(unittest.TestCase):
@mock.patch('sandbox.module.Reference')
def test_method_calls(self, mock_reference):
subject = Subject()
subject.run(1, 2, 3, x=44, y=55, z=66)
mock_reference.assert_called_with(1, 2, 3, x=44, y=55, z=66)
mock_reference.method_a.assert_called_with(1, 2, 3)
结果是
AssertionError: Expected call: method_a(1, 2, 3)
Not called
mock_reference.mock_calls
的值是
[
call(1, 2, 3, x=44, y=55, z=66),
call().method_a(1, 2, 3)
]
如果我以call().method_a
的身份访问来电,我可以访问方法详细信息,但mock_calls
会添加一个项目call()
。这可能会以一种我不希望的方式改变assert_called_once_with
,并且感觉不太对劲。此外,如果使用autospec=True
,我需要再次传递params。仅在call
中使用mock_reference.call.method_a
也不起作用。
访问mock_calls
时输出call().method_a.mock_calls
mock_reference().method_a.mock_calls
[
call(1, 2, 3, x=44, y=55, z=66),
call().method_a(1, 2, 3),
call()
]
mock_reference.assert_called_once_with(1, 2, 3)
AssertionError: Expected 'Reference' to be called once. Called 2 times.
答案 0 :(得分:13)
您模拟了一个类,但第二个方法是在实例上调用的。变化
mock_reference.method_a.assert_called_with(1, 2, 3)
到
mock_reference.return_value.method_a.assert_called_with(1, 2, 3)