python - 如何模拟 ZipFile 构造函数？

Question

我正在尝试测试此代码:

def read_classes(file):
    if CLASSES in file:
        classes = open(file, "rb").read()
    else:
        with ZipFile(file, "r") as archive:
            classes = archive.read(CLASSES)
    return classes

对我来说重要的是，当提供的文件包含 CLASSES 时以它的名字，open将被调用，否则，ZipFile将会被使用。我已经能够测试第一部分，但是，我无法模拟 ZipFile为了返回一个模拟对象( archive ) - 然后我可以断言它具有 read方法调用。这是我到目前为止一直在尝试的:

@patch('zipfile.ZipFile')
def test_givenFile_whenReadClasses_expectArchiveCalled(self, mock_zipfile):
    file = 'sample.file'
    archive = Mock()
    mock_zipfile.return_value = archive

    read_classes(file)

    archive.read.assert_called_once_with("classes.file")

当我这样做时，它会继续执行原始的 ZipFile构造函数，给我: IOError: [Errno 2] No such file or directory: 'sample.file'

Answer 1

开门见山:

@patch('zipfile.ZipFile')
def test_givenFile_whenReadClasses_expectArchiveCalled(self, mocked_zip_file):
    file = 'file'
    archive = Mock()
    mocked_read = Mock()
    archive.return_value.read = mocked_read
    mocked_zip_file.return_value.__enter__ = archive

    read_classes(dex_file)

    mocked_read.assert_called_once_with('another_file')