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Home/ Questions/Q 7072361
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T05:47:58+00:00 2026-05-28T05:47:58+00:00

Doing something like this: from zipfile import ZipFile #open zip file zipfile = ZipFile(‘Photo.zip’)

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Doing something like this:

from zipfile import ZipFile

#open zip file
zipfile = ZipFile('Photo.zip')

#iterate zip contents
for zipinfo in zipfile.filelist:
    #do something
    filepath, filename = path.split(zipinfo.filename)

How do I know if zipinfo is a file or a directory?

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  1. Editorial Team
    Editorial Team
    2026-05-28T05:47:59+00:00Added an answer on May 28, 2026 at 5:47 am

    Probably this is the right way:

    is_dir = lambda zipinfo: zipinfo.filename.endswith('/')
    
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