Python在动态文件系统中创建Zip文件

Question

I'm having trouble creating a zip in my file system which is dynamic depending on what's defined in the .env file. Can someone guide me through the correct way of writing the zip file?

Here's what I've come up so far which is I know is incorrect.

import os
import zipfile
from fs import open_fs


def download_zip_report():
    try:
        fs = open_fs(os.getenv('UPLOAD_FOLDER').strip("'"))
        with fs.open('/', 'wb') as test_file:
            zf = zipfile.ZipFile('test2.zip', mode='w')
            zf.write('file.txt')
            zf.close()

    except Exception as e:
        logging.error(str(e))
        return Response(status=404)

Answer 1

I'm not sure what the file system package is used for but the appropriate way to write a zipfile is:

from zipfile import ZipFile

with ZipFile('path/to/zipfile.zip', 'w') as zip_file:
    zip_file.write('path/to/file')

Bare in mind that "...ZipFile is also a context manager and therefore supports the with statement. In the example, myzip is closed after the with statement's suite is finished—even if an exception occurs.." (From zipfile — Work with ZIP archives documentation ) so you don't have to use zip.close() at the end. Thus applies to all context manager in Python.