Zip a directory target

Question

I'm using this basic script in Python 2.6 to zip a directory :

def zipdir(path, ziph):
    import os, zipfile
    for(dir, _, files) in os.walk(path):
        for file in files:
            ziph.write(os.path.join(dir, file))

ziph = zipfile.ZipFile(name + '.zip', 'w', zipfile.ZIP_DEFLATED)
path = 'c:/test/directory'

The problem is inside my new zip file.

For example, my target folder to zip is on c:/test/directory/myfiles

So when I zip it, instead of getting : file.zip/directory/myfiles,

I have : file.zip/test/directory/myfiles

I don't want to keep the "test" folder.

Someone could tell me how to fix it ?

You can use `shutil.make_archive` instead, use it like: `shutil.make_archive('c:/temp/compressed', 'zip', base_dir='directory', root_dir='c:/test')`. — T4rk1n, Jul 02 '18 at 15:41
It seems shutil.make_archive is not compatible with Python 2.6. — GrandGTO, Jul 03 '18 at 07:11

tstn3357 · Accepted Answer · 2018-07-03T07:58:47.963

In python 2.7 or later :

Try to use "shutil" package

import shutil
import os

zip_path = 'c:/test/new_zip_file'
target_dir_path = 'c:/test/directory'

# such as "c:/test/"
parent_target_dir_path = os.path.dirname(target_dir_path)
# such as "directory"
target_dir_name = os.path.basename(target_dir_path)

shutil.make_archive(zip_path, 'zip', root_dir=parent_target_dir_path, base_dir=target_dir_name)

The result :

new_zip_file.zip
└── directory
    ├── myfile1
    ├── myfile2
    └── myfile3 ...

EDITED

In python 2.6:

import zipfile

def zipdir(path, ziph):
    import os, zipfile
    parent_dir_name = os.path.basename(path)
    for(dir, _, files) in os.walk(path):
        for file in files:
            path_in_zip = dir[dir.find(parent_dir_name):]
            ziph.write(os.path.join(dir, file) , arcname = path_in_zip+"/"+file)

name = "ziptest"

ziph = zipfile.ZipFile(name + '.zip', 'w', zipfile.ZIP_DEFLATED)
path = 'c:/test/directory'

zipdir(path, ziph)

It uses argument arcname and specify in-zip-path.
For example, ziph.write("c:/test/directory/myfile1" , arcname = "dirctory/myfile1") makes a zipfile such as ziptest.zip/dirctory/myfile1

It seems shutil.make_archive is not available on Python 2.6 – GrandGTO Jul 03 '18 at 07:06 — GrandGTO, Jul 03 '18 at 07:06

score 0 · Answer 2 · answered Sep 11 '19 at 06:16

Python 2.7 and later:

import shutil

zip_path = 'c:/Desktop/zip_name'  # Note: no '.zip' in the suffix!
root_folder = 'c:/Documents/directory/'
shutil.make_archive(zip_path, 'zip', root_folder)

Python 2.6:

import os
import zipfile

def my_make_archive(zip_path, root_folder):
    zf = zipfile.ZipFile(zip_path, 'w')
    for (folder, sub_folder, filenames) in os.walk(root_folder):
        for filename in filenames:
            abs_file_path = os.path.join(folder, filename)
            rel_file_path = os.path.join(os.path.relpath(folder, root_folder), filename)
            zf.write(abs_file_path, arcname=rel_file_path)
    zf.close()

zip_path = 'c:/Desktop/zip_name.zip'  # Note: has '.zip' in the suffix!
root_folder = 'c:/Documents/directory/'
my_make_archive(zip_path, root_folder)

Zip a directory target

2 Answers2