Python：将文件解压缩到当前工作目录,但不保存zip中的目录结构

2024年1月21日 409次阅读

我有一个像这样的zip文件：

myArchive.zip
|
-folder1
   |
   --folder2
        |
        ---myimage.jpg

当我尝试提取myimage.jpg时：

with zipfile.ZipFile('myArchive.zip', 'r') as zfile:
   zfile.extract('folder1/folder2/myimage.jpg')

我将在我当前工作的目录中获得/folder1/folder2/myimage.jpg

但我只想将myimage.jpg提取到当前工作目录,我该怎么办呢？

最佳答案 而不是使用extract或extractall,只需获取数据并将其写入您喜欢的任何文件.这是一个代码示例,可以满足您的需求：

import os, sys, time
import zipfile

ENC = 'cp437'
outdir = unicode(os.path.abspath('.'))
outzip = 'c:/1temp/timbersales.zip'
zf = zipfile.ZipFile(outzip, 'r')


for info in zf.infolist():
    fn, dtz = info.filename, info.date_time # , info.file_size

    # some zips have dirs listed as files. Catch
    # and bypass those.
    name = os.path.basename(fn)
    if not name:
        continue

    # get our filename converted from bytes to unicode
    fn_uni = fn.decode(ENC, 'replace')
    bn_uni = os.path.basename(fn_uni)


    # this method is about 15% faster than extractall, and 
    # preserves modify and access dates
    c = zf.open(fn)
    outfile = os.path.join(outdir, bn_uni)

    # try/except to avoid problems with locked files, etc
    # do in chunks to avoid memory problems
    chunk = 2**16
    try:
        with open(outfile, 'wb') as f:
            s = c.read(chunk)
            f.write(s)
            while not len(s) < chunk: 
                s = c.read(chunk)
                f.write(s)
        c.close()
        # set modify and access dates to that inside the zip
        dtout = time.mktime(dtz + (0, 0, -1))
        os.utime(outfile, (dtout, dtout))
    except IOError:
        c.close()

此示例执行zip中的所有文件,但您可以轻松添加几行来检查特定文件.它还将覆盖工作目录中的任何文件,其名称与提取的文件同名.

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