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

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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'))
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rcbevans
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rcbevans

I'm a graduate from Imperial College London in Electronics and Computer Science, ex Software Engineer at ARM, and now a Software Engineer at Microsoft in Redmond, WA. I'm a lover of all things tech and/or powered by petrol, especially motorcycles.

Updated on September 10, 2020

Comments

  • rcbevans
    rcbevans over 3 years

    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?