如何在不维护Python目录结构的情况下从zip中提取文件？ [英] How to extract file from zip without maintaining directory structure in Python?

问题描述

我正尝试使用python从zip存档中提取特定文件。

I'm trying to extract a specific file from a zip archive using python.

在这种情况下，请从apk本身提取apk的图标。

In this case, extract an apk's icon from the apk itself.

我当前正在使用

with zipfile.ZipFile('/path/to/my_file.apk') as z:
    # extract /res/drawable/icon.png from apk to /temp/...
    z.extract('/res/drawable/icon.png', 'temp/')

这确实有效，在我的脚本目录中正在创建 temp / res / drawable / icon.png 是temp加上与文件在apk内相同的路径。

which does work, in my script directory it's creating temp/res/drawable/icon.png which is temp plus the same path as the file is inside the apk.

我实际上想要得到的最终结果是 temp / icon.png 。

What I actually want is to end up with temp/icon.png.

有没有办法直接使用zip命令，还是我需要解压缩，然后移动文件，然后手动删除目录？

Is there any way of doing this directly with a zip command, or do I need to extract, then move the file, then remove the directories manually?

推荐答案

您可以使用< a href = http://docs.python.org/3/library/zipfile.html#zipfile.ZipFile.open rel = nofollow noreferrer> zipfile.ZipFile.ope n ：

import shutil
import zipfile

with zipfile.ZipFile('/path/to/my_file.apk') as z:
    with z.open('/res/drawable/icon.png') as zf, open('temp/icon.png', 'wb') as f:
        shutil.copyfileobj(zf, f)

或使用 zipfile.ZipFile.read ：

import zipfile

with zipfile.ZipFile('/path/to/my_file.apk') as z:
    with open('temp/icon.png', 'wb') as f:
        f.write(z.read('/res/drawable/icon.png'))

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