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python-3.x - 如何断言在 pytest 中调用了猴子补丁？

问题描述

考虑以下：

class MockResponse:
    status_code = 200

    @staticmethod
    def json():
        return {'key': 'value'}
                                  # where api_session is a fixture
def test_api_session_get(monkeypatch, api_session) -> None:
    def mock_get(*args, **kwargs):
        return MockResponse()

    monkeypatch.setattr(requests.Session, 'get', mock_get)
    response = api_session.get('endpoint/') # My wrapper around requests.Session
    assert response.status_code == 200
    assert response.json() == {'key': 'value'}
    monkeypatch.assert_called_with(
        'endpoint/',
        headers={
            'user-agent': 'blah',
        },
    )

我如何断言get我正在修补的被调用'/endpoint'and headers？当我现在运行测试时，我收到以下失败消息：

FAILED test/utility/test_api_session.py::test_api_session_get - AttributeError: 'MonkeyPatch' object has no attribute 'assert_called_with'

我在这里做错了什么？感谢所有提前回复的人。

标签： python-3.xpython-requestsmockingpytestmonkeypatching

解决方案

将添加另一个使用monkeypatch 而不是“你不能使用monkeypatch”的响应

由于python有闭包，这是我可怜的人用monkeypatch做这些事情的方式：

patch_called = False

def _fake_delete(keyname):
    nonlocal patch_called
    patch_called = True
    assert ...

monkeypatch.setattr("mymodule._delete", _fake_delete)
res = client.delete(f"/.../{delmeid}"). # this is a flask client
assert res.status_code == 200
assert patch_called

在您的情况下，由于我们正在通过修补 HTTP 服务器方法处理程序来做类似的事情，您可以做类似的事情（不是说这很漂亮）：

param_called = None

def _fake_delete(param):
    nonlocal param_called
    patch_called = param
    assert ...

monkeypatch.setattr("mymodule._delete", _fake_delete)
res = client.delete(f"/.../{delmeid}")
assert res.status_code == 200
assert param_called == "whatever this should be"

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