How do I zip the contents of a folder using python (version 2.5)?

Question

Once I have all the files I require in a particular folder, I would like my python script to zip the folder contents.

Is this possible?

And how could I go about doing it?

Answer 1

On python 2.7 you might use: shutil.make_archive(base_name, format[, root_dir[, base_dir[, verbose[, dry_run[, owner[, group[, logger]]]]]]]).

base_name archive name minus extension

format format of the archive

root_dir directory to compress.

For example

 shutil.make_archive(target_file, format="bztar", root_dir=compress_me)    

Answer 2

Adapted version of the script is:

#!/usr/bin/env python
from __future__ import with_statement
from contextlib import closing
from zipfile import ZipFile, ZIP_DEFLATED
import os

def zipdir(basedir, archivename):
    assert os.path.isdir(basedir)
    with closing(ZipFile(archivename, "w", ZIP_DEFLATED)) as z:
        for root, dirs, files in os.walk(basedir):
            #NOTE: ignore empty directories
            for fn in files:
                absfn = os.path.join(root, fn)
                zfn = absfn[len(basedir)+len(os.sep):] #XXX: relative path
                z.write(absfn, zfn)

if __name__ == '__main__':
    import sys
    basedir = sys.argv[1]
    archivename = sys.argv[2]
    zipdir(basedir, archivename)

Example:

C:\zipdir> python -mzipdir c:\tmp\test test.zip

It creates 'C:\zipdir\test.zip' archive with the contents of the 'c:\tmp\test' directory.

Answer 3

Here is a recursive version

def zipfolder(path, relname, archive):
    paths = os.listdir(path)
    for p in paths:
        p1 = os.path.join(path, p) 
        p2 = os.path.join(relname, p)
        if os.path.isdir(p1): 
            zipfolder(p1, p2, archive)
        else:
            archive.write(p1, p2) 

def create_zip(path, relname, archname):
    archive = zipfile.ZipFile(archname, "w", zipfile.ZIP_DEFLATED)
    if os.path.isdir(path):
        zipfolder(path, relname, archive)
    else:
        archive.write(path, relname)
    archive.close()

Answer 4

Both jfs's solution and Kozyarchuk's solution could work for the OP's use case, however:

• jfs's solution zips all of the files in a source folder and stores them in the zip at the root level (not preserving the original source folder within the structure of the zip).
• Kozyarchuk's solution inadvertently puts the newly-created zip file into itself since it is a recursive solution (e.g. creating new zip file "myzip.zip" with this code will result in the archive "myzip.zip" itself containing an empty file "myzip.zip")

Thus, here is a solution that will simply add a source folder (and any subfolders to any depth) to a zip archive. This is motivated by the fact that you cannot pass a folder name to the built-in method ZipFile.write() -- the function below, add_folder_to_zip(), offers a simple method to add a folder and all of its contents to a zip archive. Below code works for Python2 and Python3.

import zipfile
import os

def add_folder_to_zip(src_folder_name, dst_zip_archive):
    """ Adds a folder and its contents to a zip archive

        Args:
            src_folder_name (str): Source folder name to add to the archive
            dst_zip_archive (ZipFile):  Destination zip archive

        Returns:
            None
    """
    for walk_item in os.walk(src_folder_name):
        for file_item in walk_item[2]:
            # walk_item[2] is a list of files in the folder entry
            # walk_item[0] is the folder entry full path 
            fn_to_add = os.path.join(walk_item[0], file_item)
            dst_zip_archive.write(fn_to_add)

if __name__ == '__main__':
    zf = zipfile.ZipFile('myzip.zip', mode='w')
    add_folder_to_zip('zip_this_folder', zf)
    zf.close()