从zip中提取文件而不使用python ZipFile保留结构？

Question

我尝试从包含一个文件夹中的子文件夹的.zip中提取所有文件.我希望子文件夹中的所有文件只在一个文件夹中提取,而不保留原始结构.此刻,我提取所有内容,将文件移动到文件夹,然后删除以前的子文件夹.具有相同名称的文件被覆盖.

在写文件之前可以这样做吗？

这是一个结构,例如:

my_zip/file1.txt
my_zip/dir1/file2.txt
my_zip/dir1/dir2/file3.txt
my_zip/dir3/file4.txt

最后我告诉你:

my_dir/file1.txt
my_dir/file2.txt
my_dir/file3.txt
my_dir/file4.txt

我可以在此代码中添加什么内容？

import zipfile
my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"

zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
    zip_file.extract(files, my_dir)
zip_file.close()

如果我从zip_file.namelist()重命名文件路径,我有这个错误:

KeyError: "There is no item named 'file2.txt' in the archive"

Answer 1

这将打开zip存档成员的文件句柄,提取文件名并将其复制到目标文件(这是如何ZipFile.extract工作的,不需要处理子目录).

import os
import shutil
import zipfile

my_dir = r"D:\Download"
my_zip = r"D:\Download\my_file.zip"

with zipfile.ZipFile(my_zip) as zip_file:
    for member in zip_file.namelist():
        filename = os.path.basename(member)
        # skip directories
        if not filename:
            continue

        # copy file (taken from zipfile's extract)
        source = zip_file.open(member)
        target = open(os.path.join(my_dir, filename), "wb")
        with source, target:
            shutil.copyfileobj(source, target)

Answer 2

有可能迭代ZipFile.infolist().在返回的ZipInfo对象上,您可以操作filename删除目录部分,最后将其提取到指定的目录.

import glob
import zipfile
import shutil
import os

my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"

with zipfile.ZipFile(my_zip) as zip:
    for zip_info in zip.infolist():
        if zip_info.filename[-1] == '/':
            continue
        zip_info.filename = os.path.basename(zip_info.filename)
        zip.extract(zip_info, my_dir)

Answer 3

只需提取到内存中的字节,计算文件名,然后自己编写,而不是让库做 - 只是使用"read()"而不是"extract()"方法:

import zipfile
import os

my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"

zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
    data = zip_file.read(files, my_dir)
    # I am almost shure zip represents directory separator
    # char as "/" regardless of OS, but I  don't have DOS or Windos here to test it
    myfile_path = os.path.join(my_dir, files.split("/")[-1])
    myfile = open(myfile_path, "wb")
    myfile.write(data)
    myfile.close()
zip_file.close()

Answer 4

与Gerhard Götz 的解决方案类似的概念，但适用于提取单个文件而不是整个 zip：

with ZipFile(zipPath, 'r') as zipObj:
    zipInfo = zipObj.getinfo(path_in_zip))
    zipInfo.filename = os.path.basename(destination)
    zipObj.extract(zipInfo, os.path.dirname(os.path.realpath(destination)))