How to extract file from zip without maintaining directory structure in Python?
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'))
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, 2020Comments
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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?