Python;如何在不包含父目录的情况下提取文件

Question

我是 Python 的菜鸟，我需要帮助！我正在尝试将 zipfile 模块用于最新版本的 Python 3，并且在准备提取 zip 文件时尝试排除父目录（文件夹）时遇到了麻烦。

zip 文件目录的示例如下所示：

Content/
Content/Runtime/
Content/Runtime/test.txt
Content/Runtime/test.png
Content/Documentation/
Content/Documentation/license.txt

我不知道如何从提取中排除父目录（第一个文件夹）。我只想提取运行时和文档目录及其内容。

请帮忙，谢谢，非常感谢！

Answer 1

在这种情况下，用于提取的内置方法将不起作用（因为它们会提取完整路径），但您可以自己轻松完成：

import os
import shutil

destination_dir = './destination'

with zipfile.ZipFile('./tmp.zip', 'r') as z:
    # See https://docs.python.org/3/library/zipfile.html#zipfile.ZipFile.infolist
    for file_info in z.infolist():
        # Only extract regular files
        if file_info.is_dir():
            continue

        file_path = file_info.filename
        # Only extract things under 'Content/'
        if not file_path.startswith('Content/'):
            continue

        # Split at slashes, at most one time, and take the second part
        # so that we skip the 'Content/' part
        extracted_path = file_path.split('/', 1)[1]

        # Combine with the destination directory
        extracted_path = os.path.join(destination_dir, extracted_path)
        print(extracted_path)

        # Make sure the directory for the file exists
        os.makedirs(os.path.dirname(extracted_path), exist_ok=True)

        # Extract the file. Don't use `z.extract` as it will concatenate
        # the full path from inside the zip.
        # WARNING: This code does not check for path traversal vulnerabilities
        # Refer to the big warning inside the ZipFile module for more details
        with open(extracted_path, 'wb') as dst:
          with z.open(file_info, 'r') as src:
            shutil.copyfileobj(src, dst)