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

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)    

Adapted version of the script is:

#!/usr/bin/env pythonfrom __future__ import with_statementfrom contextlib import closingfrom zipfile import ZipFile, ZIP_DEFLATEDimport osdef 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.

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()