How do I create a zip file of a file path using Python, including empty directories?

Question

I've been trying to use the zipfile and shutil.make_archive modules to recursively create a zip file of a directory. Both modules work great--except empty directories do not get added to the archive. Empty directories containing other empty directories are also silently skipped.

I can use 7Zip to create an archive of the same path and empty directories are preserved. Therefore I know this is possible within the file format itself. I just don't know how to do it within Python. Any ideas? Thanks!

Answer 1

There is a example using zipfile:

import os, zipfile  
from os.path import join  
def zipfolder(foldername, filename, includeEmptyDIr=True):   
    empty_dirs = []  
    zip = zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED)  
    for root, dirs, files in os.walk(foldername):  
        empty_dirs.extend([dir for dir in dirs if os.listdir(join(root, dir)) == []])  
        for name in files:  
            zip.write(join(root ,name))  
        if includeEmptyDIr:  
            for dir in empty_dirs:  
                zif = zipfile.ZipInfo(join(root, dir) + "/")  
                zip.writestr(zif, "")  
        empty_dirs = []  
    zip.close() 

if __name__ == "__main__":
    zipfolder('test1/noname/', 'zip.zip')

Answer 2

You'll need to register a new archive format to do that, since the default ZIP archiver does not support that. Take a look at the meat of the existing ZIP archiver. Make your own archiver that creates directories using that currently-unused dirpath variable. I looked for how to create an empty directory and found this:

zip.writestr(zipfile.ZipInfo('empty/'), '')

With that, you should be able to write the necessary code to make it archive empty directories.