How to extract file from zip without maintaining directory structure in Python?

rcbevans picture rcbevans · Jul 18, 2013 · Viewed 24.6k times · Source

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

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

I am currently using

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.png which is temp plus the same path as the file is inside the apk.

What I actually want is to end up with 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?

Answer

falsetru picture falsetru · Jul 18, 2013

You can use zipfile.ZipFile.open:

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)

Or use 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'))