如何在Python中压缩文件夹及其所有文件，同时保留其内容的文件夹名称和相对路径？

Question

I am trying to create a backup script in Python 3.5 that will zip all the contents of a specified folder. Here is what I am doing now:

import shutil
import datetime
import os
import os.path

import dropbox


INPUT_FOLDER = '~/Documents/db-backup-test/'
ARCHIVE_PATH = '~/Documents/tmp/'
ARCHIVE_FILE_NAME = "db-backup_iOS__"


def compress_folder(origem, destino):
    try:
        arq = shutil.make_archive(destino, 'zip', root_dir=origem, base_dir=origem)
    except Exception as e:
        print('\n\nAn error was found during compression:')
        print(e)


timestamp = str(datetime.datetime.now())
input_path = os.path.expanduser(INPUT_FOLDER)
archive_path = os.path.expanduser(ARCHIVE_PATH + ARCHIVE_FILE_NAME + timestamp)

compress_folder(input_path, archive_path)

I goes well, except for one thing: the resulting zip archive extracts its contents with a reproduction of the full original path. What I would like is to get is a simple copy of the folder, with its original name and its contents with relative paths, when extracting the archive.

I was looking into a simple and short solution that would not require to iterate through the folder contents.

Is this possible with shutil.make_archive() , or will I need to dig into other modules, like zipfile ?

Answer 1

Use the arcname parameter to control the name/path in the zip file.

#!/usr/bin/env python2.7

import os
import zipfile

def zip(src, dst):
   zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
   abs_src = os.path.abspath(src)
   for dirname, subdirs, files in os.walk(src):
       for filename in files:
          absname = os.path.abspath(os.path.join(dirname, filename))
          arcname = absname[len(abs_src) + 1:]
          print 'zipping %s as %s' % (os.path.join(dirname, filename), arcname)
          zf.write(absname, arcname)
   zf.close()
zip("src", "dst")

With a directory structure like this:

src
└── a
    ├── b
    │   └── bar
    └── foo