嘲笑不使用pytest和flexmock

Question

我正在尝试使用pytest fixture来模拟调用open(),然后在测试拆卸时重置它,但由于某种原因,mock不会应用于测试函数中.

这是我的样本:

# tests.py
@pytest.fixture(scope='module')
def mock_open(request):
    mock = flexmock(sys.modules['__builtin__'])
    mock.should_call('open')
    m = (mock.should_receive('open')
        .with_args('/tmp/somefile')
        .and_return(read=lambda: 'file contents'))
    request.addfinalizer(lambda: m.reset())

def test_something(mock_open):
    ret = function_to_test()
    ret[1]()  # This actually uses the original open()

而且,如果它很重要,这就是function_to_test()看起来像:

# some_module.py
def function_to_test():
    def inner_func():
        open('/tmp/somefile').read()   # <-- I want this call mocked
        # More code ...
    return (True, inner_func)

如果我使用xUnit-style setup_module()/ teardown_module()functions 也会发生这种情况.但是如果我把模拟代码放在测试函数本身(我显然不想这样做),那么它工作正常.

我错过了什么？谢谢!

Answer 1

怎么用mock？

tests.py:

import mock # import unittest.mock (Python 3.3+)
import pytest

from some_module import function_to_test

@pytest.fixture(scope='function')
def mock_open(request):
    m = mock.patch('__builtin__.open', mock.mock_open(read_data='file content'))
    m.start()
    request.addfinalizer(m.stop)

def test_something(mock_open):
    ret = function_to_test()
    assert ret[1]() == 'file content'

some_module.py:

def function_to_test():
    def inner_func():
        return open('/tmp/somefile').read()
    return True, inner_func