Better way to zip files in Python (zip a whole directory with a single command)? [duplicate]

This question already has answers here: Closed 10 years ago.

Possible Duplicate:

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

Suppose I have a directory: /home/user/files/. This dir has a bunch of files:

/home/user/files/
  -- test.py
  -- config.py

I want to zip this directory using ZipFile in python. Do I need to loop through the directory and add these files recursively, or is it po开发者_JAVA百科ssible do pass the directory name and the ZipFile class automatically adds everything beneath it?

In the end, I would like to have:

/home/user/files.zip (and inside my zip, I dont need to have a /files folder inside the zip:)
  -- test.py
  -- config.py

---

Note that this doesn't include empty directories. If those are required there are workarounds available on the web; probably best to get the ZipInfo record for empty directories in our favorite archiving programs to see what's in them.

Hardcoding file/path to get rid of specifics of my code...

target_dir = '/tmp/zip_me_up'
zip = zipfile.ZipFile('/tmp/example.zip', 'w', zipfile.ZIP_DEFLATED)
rootlen = len(target_dir) + 1
for base, dirs, files in os.walk(target_dir):
   for file in files:
      fn = os.path.join(base, file)
      zip.write(fn, fn[rootlen:])

---

You could try using the distutils package:

distutils.archive_util.make_zipfile(base_name, base_dir[, verbose=0, dry_run=0])

---

You might also be able to get away with using the zip command available in the Unix shell with a call to os.system

---

You could use subprocess module:

import subprocess

PIPE = subprocess.PIPE
pd = subprocess.Popen(['/usr/bin/zip', '-r', 'files', 'files'],
                      stdout=PIPE, stderr=PIPE)
stdout, stderr = pd.communicate()

The code is not tested and pretends to works just on unix machines, i don't know if windows has similar command line utilities.