如何在Pytest中mock任意路径的文件写入操作，避免真实生成文件

你可以直接使用Python标准库unittest.mock中的patch和mock_open组合实现需求，不需要触发真实的文件IO操作：

基础实现（单文件写入场景） #

from unittest.mock import patch, mock_open

# 替换为你自己业务模块的导入，比如你的写文件逻辑在utils.py的save_binary函数中
from your_module import save_binary

def test_file_write_logic():
    # 初始化mock的open对象
    mocked_open = mock_open()

    # patch全局内置open函数，替换为mock对象
    with patch("builtins.open", mocked_open):
        test_path = "/any/non/exist/path.bin"
        test_byte_data = [b"part1", b"part2", b"part3"]
        
        # 执行业务逻辑
        save_binary(test_path, test_byte_data)

        # 验证open调用参数符合预期
        mocked_open.assert_called_once_with(test_path, "wb")
        # 验证写入内容符合预期
        file_handle = mocked_open()
        expected_content = b"".join(test_byte_data)
        file_handle.write.assert_called_once_with(expected_content)

多文件写入场景实现 #

如果业务逻辑中会打开多个不同路径的文件，你可以给mock对象配置side_effect来区分不同路径的调用：

from unittest.mock import patch, mock_open, call

from your_module import multi_file_write_logic

def test_multi_file_write():
    # 存储不同路径对应的mock文件句柄
    path_handle_map = {}
    def open_side_effect(file_path, *args, **kwargs):
        if file_path not in path_handle_map:
            path_handle_map[file_path] = mock_open()()
        return path_handle_map[file_path]

    with patch("builtins.open", side_effect=open_side_effect) as mocked_open:
        # 执行业务逻辑
        multi_file_write_logic()

        # 验证指定路径的open调用
        target_path = "/target/file/path"
        mocked_open.assert_has_calls([call(target_path, "wb")], any_order=True)
        # 验证指定路径的写入内容
        target_handle = path_handle_map[target_path]
        target_handle.write.assert_called_once_with(b"expected content")

注意：mock失效的绝大多数原因是patch路径错误，你需要保证patch的open路径是业务代码实际引用的路径，比如业务代码中用了from builtins import open as custom_open，你需要patch业务模块下的custom_open，而不是全局的builtins.open

内容的提问来源于stack exchange，提问作者James