从 zip 存档中提取特定文件，而无需在 python 中维护目录结构
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Extract a specific file from a zip archive without maintaining directory structure in python
提问by rcbevans
I'm trying to extract a specific file from a zip archive using python.
我正在尝试使用 python 从 zip 存档中提取特定文件。
In this case, extract an apk's icon from the apk itself.
在这种情况下，从 apk 本身提取 apk 的图标。
I am currently using
我目前正在使用
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/')
which does work, in my script directory it's creating temp/res/drawable/icon.pngwhich is temp plus the same path as the file is inside the apk.
这确实有效，在我的脚本目录中，它正在创建temp/res/drawable/icon.png临时加上与文件在 apk 内的路径相同的路径。
What I actually want is to end up with temp/icon.png.
我真正想要的是以temp/icon.png.
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?
有没有办法直接使用 zip 命令执行此操作，或者我是否需要提取，然后移动文件，然后手动删除目录？
采纳答案by falsetru
You can use zipfile.ZipFile.open:
您可以使用zipfile.ZipFile.open：
import os
import shutil
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)
Or use zipfile.ZipFile.read:
或者使用zipfile.ZipFile.read：
import os
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'))
