使用python提取zip文件并保留顶级文件夹 [英] Extract zip file and keeping top folder using python

问题描述

我有像 CW1234.zip 这样的文件夹，它有各种文件夹和子文件夹，如下所示.所以，CW1234.zip 有 CW_All 文件夹，而后有 CW123 和 CW234 文件夹等等

I have folder in like CW1234.zip and it has various folders and subfolders like below. So, CW1234.zip has CW_All folder which in turn has CW123 and CW234 folders and so on

CW1234.zip
  CW_All
    CW123
      xyz.pdf
    CW234
      abc.doc

并提取我使用此代码:

from zipfile import ZipFile

with ZipFile(r'CW41234.zip', 'r') as zipObj:
   # Extract all the contents of zip file in current directory
   zipObj.extract()

唯一的问题是我得到的解压文件夹来自 CW_All 以及所有子文件夹和文件.

The only problem is the unzipped folder I get is from CW_All and all the subfolders and file.

我想要的是从 CW1234 中获取它作为一个文件夹，然后结构如下?

What I want is to get it from CW1234 as one folder and then the structure follows?

电流输出

CW_All
   CW123
      xyz.pdf
   CW234
      abc.doc

预期产出

CW1234
  CW_All
     CW123
        xyz.pdf
     CW234
        abc.doc

在文档中也找不到任何东西！！

Couldn't find anything in the documentation also!!

推荐答案

使用 ZipFile.extractall() 我们可以简单地提供一个新的路径来提取档案的内容，我们可以根据档案的文件名.

Using ZipFile.extractall() we can simply provide a new path to extract the contents of the archive to, which we can base on the filename of the archive.

我有一个具有以下结构的 .zip 文件:

I have a .zip file with the following structure:

archive1024.zip:.
│
└───Folder_with_script
        stuff.py

这是将存档中的所有文件提取到子文件夹中的脚本:

Here is the script to extract all of the files inside of the archive into a sub-folder:

from zipfile import ZipFile

file = "archive1024.zip"
with ZipFile(file, "r") as zFile:
    zFile.extractall(path=file.split(".")[0])

我现在有一个这样的文件夹结构:

I now have a folder-structure like this:

J:.
│   archive1024.zip
│   unzip.py
│
└───archive1024
    └───Folder_with_script
            stuff.py

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