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

最后我希望如此：

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（这就是ZipFile.extract的工作方式，而不需要处理子目录）。

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

可以迭代ZipFile.infolist()。在返回的ZipInfo对象上，您可以操作filename以删除目录部分，最后将其提取到指定目录。

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the solution of Gerhard Götz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)))

Answer 5

如果你收到badZipFile错误。您可以使用7zip子进程解压缩存档。假设您已安装7zip，请使用以下代码。

import subprocess
my_dir = destFolder #destination folder
my_zip = destFolder + "/" + filename.zip #file you want to extract
ziploc = "C:/Program Files/7-Zip/7z.exe" #location where 7zip is installed
cmd = [ziploc, 'e',my_zip ,'-o'+ my_dir ,'*.txt' ,'-r' ] 
#extracting only txt files and from all subdirectories
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)