Python 我想模拟打开一个现有的测试文件，为测试应用一个函数，但不更改文件的实际内容

我有一个更新rpm规范文件以供以后使用的函数。它有三个参数：应用程序的名称、我们想要升级到的版本以及spec文件在文件系统中的位置

我似乎不能做的是，比如说，拥有一个tests/files/package_name.spec文件，并测试它是否在不更改实际spec文件的情况下正确地对其进行了版本设置。我不想更改实际的spec文件，因为它会使测试在第二次运行时变得无用

我试着查看模拟图书馆和模拟开放，但似乎没有什么适合这种情况

def test_update_spec_file():
    update_spec_file("package_name", "5.0.0", "files/package_name")
    with open("files/package_name.spec", "r") as f:
        contents = f.read()
        version_regex = re.compile("^.*define _software_version.*$")
        assert (
            re.search(version_regex, contents).group(0)
            == "%define _software_version 5.0.0"
        )

基本上，

update\u spec\u file

函数所做的是使用regex查找

%define\u software\u version 5.0.0

行，并再次使用regex在新版本中进行sub

总之，是否有可能让文件存在，用某种模拟文件及其内容打开它，在该文件上运行函数，并断言它完成了它的工作，而不实际修改文件系统中的文件

或者在修改文件后重置该文件

以下是更新等级库文件的代码：

def update_spec_file(application, release_version, path_to_spec_file):
    """Update the spec file with the release version.

    Args:
        application (str): Name of the application.
        release_version (str): New version of application.
        path_to_spec_file (str, optional):
            path the the applications spec file we are modifying.
    """
    print(f"Updating spec file _software_version to {release_version}")

    try:
        with open(path_to_spec_file, "r") as spec_file:
            spec_contents = spec_file.readlines()
    except EnvironmentError:
        print(f"Unable to open spec file located at: {path_to_spec_file}")
        # raise the original exception again.
        raise

    # Find and replace software version in contents of spec file.
    version_regex = re.compile("%define _software_version.*?(?=(?:\\n)|$)")
    define_version = f"%define _software_version {release_version}"
    # Splice the new software version line into the existing file contents
    spec_contents[:] = [
        version_regex.sub(define_version, line) for line in spec_contents
    ]

    try:  # Open the file and write the new spliced contents
        with open(path_to_spec_file, "w") as f_spec_file:
            f_spec_file.writelines(spec_contents)
    except EnvironmentError:  # Catch spec file not found.
        print(f"Unable to open spec file located at: {path_to_spec_file}")
        # raise the original exception again.
        raise

编辑：

这是修改后的代码：

def test_update_spec_file():
    content = open("tests/files/package_name.spec").read()
    with patch(__name__ + ".open", create=True) as mock_open:
        mock_open.return_value.__enter__.return_value = StringIO(content)
        with open("tests/files/package_name.spec") as f:
            update_spec_file(
                "package_name", "5.0.0", "tests/files/package_name.spec"

我要做的是在测试运行之前测试规范文件，在测试运行之后测试规范文件

代码的树结构：

release/
├── release_all_in_one
│   ├── changelog_utils.py
│   ├── confluence_email.py
│   ├── confluence_utils.py
│   ├── create_confluence_page.py
│   ├── __init__.py
│   ├── jira_utils.py
│   ├── parse_confluence.py
│   ├── parse_rm_objects.py
│   ├── release_all_in_one_openshift.py
│   ├── release_all_in_one.py
│   ├── tag_branch.py
│   ├── templates
│   │   ├── admin_package_template
│   │   ├── base_template
│   │   ├── cdn_package_template
│   │   ├── fe_package_template
│   │   ├── oddjob_package_template
│   │   ├── openshift_template
│   │   └── procedure.json
│   └── utils.py
├── setup.cfg
├── setup.py
└── tests
    ├── files
    │   └── package_name.spec
    ├── test_changelog.py
    ├── testing_utils.py
    └── test_utils.py

---

您可以

patch

使用

open

函数返回上下文管理器，该上下文管理器返回带有实际文件内容的

io.StringIO

对象：

from unittest.mock import patch
from io import StringIO
content = '123' # value for demo
# uncomment the line below to read actual file content
# content = open('files/package_name.spec').read()
with patch(__name__ + '.open', create=True) as mock_open:
    mock_open.return_value.__enter__.return_value = StringIO(content)
    with open('files/package_name.spec') as f:
        print(f.read())
        f.seek(0)
        f.write('abc')
        f.seek(0)
        print(f.read())

这将产生：

123
abc

或者，您可以将

open

指定给

MagicMock

对象，以便覆盖同一模块中对

open

的所有引用：

from unittest.mock import MagicMock
from io import StringIO
content = '123' # value for demo
# uncomment the line below to read actual file content
# content = open('files/package_name.spec').read()
open = MagicMock()
open.return_value.__enter__.return_value = StringIO(content)
with open('files/package_name.spec') as f:
    print(f.read())
    f.seek(0)
    f.write('abc')
    f.seek(0)
    print(f.read())

---

嘿，我很感谢您的帮助，但是这仍然会修改实际文件：\n请使用实际代码更新您的问题，以更新文件。当前发布的代码根本不修改文件。在其中添加了函数。我将尝试。它在同一模块中。我可以很快发布树结构。好的，这确实有效。谢谢你一直以来的帮助。