如何递归提取zip文件? [英] How to extract zip file recursively?

问题描述

我有一个 zip 文件，其中包含三个 zip 文件，如下所示:

I have a zip file which contains three zip files in it like this:

zipfile.zip\  
    dirA.zip\
         a  
    dirB.zip\
         b  
    dirC.zip\
         c

我想在具有这些名称(dirA、dirB、dirC)的目录中提取 zip 文件中的所有内部 zip 文件.
基本上，我想最终得到以下架构:

I want to extract all the inner zip files that are inside the zip file in directories with these names (dirA, dirB, dirC).
Basically, I want to end up with the following schema:

output\  
    dirA\
         a  
    dirB\
         b  
    dirC\
         c

我尝试了以下方法:

import os, re
from zipfile import ZipFile

os.makedirs(directory)  # where directory is "\output"
with ZipFile(self.archive_name, "r") as archive:
    for id, files in data.items():
        if files:
            print("Creating", id)
            dirpath = os.path.join(directory, id)

            os.mkdir(dirpath)

            for file in files:
                match = pattern.match(filename)
                new = match.group(2)
                new_filename = os.path.join(dirpath, new)

                content = archive.open(file).read()
            with open(new_filename, "wb") as outfile:
                outfile.write(content)

但它只提取 zip 文件，我最终得到:

But it only extracts the zip file and I end up with:

output\  
    dirA\
         dirA.zip 
    dirB\
         dirB.zip 
    dirC\
         dirC.zip

任何包括代码段的建议都将不胜感激，因为我尝试了很多不同的东西，但没有成功阅读文档.

Any suggestions including code-segments will be much appreciated cause I have tried so many different things and read the docs without success.

推荐答案

在提取 zip 文件时，您可能希望将内部 zip 文件写入内存而不是磁盘上.为此，我使用了 BytesIO.

When extracting the zip file, you would want to write the inner zip files to memory instead of them on disk. To do this, I've used BytesIO.

查看此代码:

import os
import io
import zipfile

def extract(filename):
    z = zipfile.ZipFile(filename)
    for f in z.namelist():
        # get directory name from file
        dirname = os.path.splitext(f)[0]  
        # create new directory
        os.mkdir(dirname)  
        # read inner zip file into bytes buffer 
        content = io.BytesIO(z.read(f))
        zip_file = zipfile.ZipFile(content)
        for i in zip_file.namelist():
            zip_file.extract(i, dirname)

如果您使用 zipfile.zip 运行 extract("zipfile.zip") 为:

If you run extract("zipfile.zip") with zipfile.zip as:

zipfile.zip/
    dirA.zip/
        a
    dirB.zip/
        b
    dirC.zip/
        c

输出应该是:

dirA/
  a
dirB/
  b
dirC/
  c

这篇关于如何递归提取zip文件?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！