如何压缩文件夹的内容而不是文件夹

Question

我正在使用以下文件夹结构：

.
+-- temp
|   +-- sample1.csv
|   +-- sample2.csv
+-- zips
|   +-- zip1.zip
|   +-- zip2.zip
+-- main.py

我想将 temp 文件夹的内容压缩到一个 zip 文件中，但我不想包含该文件夹。我尝试了以下方法（希望只压缩文件）：

import pathlib
import zipfile

origin = pathlib.Path('.') / 'temp'
destination = pathlib.Path('.') / 'zips' / 'zip3.zip'

with zipfile.ZipFile(destination, mode='w') as zipfile_:
    for content in origin.iterdir():
        zipfile_.write(content)

我也尝试过使用 shutil.make_archive()，但结果相同，唯一的区别是目录被简单地命名为 "."。

编辑：谢谢 CookieReaver and metatoaster (I believe) for pointing out that ZipFile.write() method has an argument called arcname that will name the file (ZipFile.write() documentation)。

这是我应用修复后的原始问题：

import pathlib
import zipfile

origin = pathlib.Path('.') / 'temp'
destination = pathlib.Path('.') / 'zips' / 'zip3.zip'

with zipfile.ZipFile(destination, mode='w') as zipfile_:
    for content in origin.iterdir():
        zipfile_.write(content, content.name)

Answer 1

import zipfile
import os

orig = r"path_to_origin_dir"
dest = r"path_to_destination_dir"

zip_name = "zip3.zip"
target_zip = zipfile.ZipFile(os.path.join(dest, zip_name), 'w')

for dp, dn, fn in os.walk(orig):

    for filename in fn:
        if filename.endswith('csv'):
            target_zip.write(os.path.join(dp, filename), os.path.relpath(os.path.join(dp, filename), orig))

target_zip.close()

这是你的目标吗？

如何压缩文件夹的内容而不是文件夹

How to zip contents of but not the folder

python

directory

zip

file

pathlib