在python中选择性提取和打开zipfile [英] Selective extracting and opening for zipfile in python

问题描述

从文档中看来，可以使用本地python 使用

From the docs, it looks like it's possible to perform selective file extract and open using the zipfile module in native python, http://docs.python.org/2/library/zipfile using

ZipFile.extract(member [，path [，pwd]])

ZipFile.extract(member[, path[, pwd]])

从存档中将成员提取到当前工作目录；成员必须是其全名或ZipInfo对象).它的档案信息将尽可能准确地提取出来.路径指定一个提取到不同的目录.成员可以是文件名或ZipInfo对象.pwd是用于加密文件的密码.

Extract a member from the archive to the current working directory; member must be its full name or a ZipInfo object). Its file information is extracted as accurately as possible. path specifies a different directory to extract to. member can be a filename or a ZipInfo object. pwd is the password used for encrypted files.

我有一个zipfile，例如 foobar.zip :

I have a zipfile as such foobar.zip:

foobar.zip\
  \foo
      \a.txt
      \b.txt
  \bar
      \b.txt
      \c.txt

我尝试从.zip文件的单个子目录中提取文件，但有时不打印任何内容:

I've tried to extract files from a single sub-directory of the .zip file but it prints nothing sometimes:

import zipfile
with zipfile.ZipFile('foobar.zip','r') as inzipfile:
  for infile in inzipfile.namelist():
    if 'foo' in os.path.split(infile)[0]:
      print inzipfile.open(infile,'r').read()

我试图给出一些可能要提取的选定文件的列表，但有时也什么也不打印.

I've tried to give a list of selected files that i might want to extract but it also prints nothing sometimes too.

wanted = ['a.txt', 'b.txt']
import zipfile
with zipfile.ZipFile('foobar.zip','r') as inzipfile:
  for infile in inzipfile.namelist():
    if os.path.split(infile)[1] in wanted:
      print inzipfile.open(infile,'r').read()

已代码或我读取文件的方式都没错.我认为我的zip文件有问题，导致 schroedinbug 有时在我的子目录中目录文件无法打开，并且 inzipfile.open(infile，'r').read()返回None.现在，它将提取，打开并打印文件的内容.

Edited: There's nothing wrong with the code or how I'm reading the files. I think there's something wrong with my zipfile which causes schroedinbug where sometimes my sub-directory files don't open and inzipfile.open(infile,'r').read() returns None. Now it extracts, opens and print the content of the file.

是否知道如何在python代码中检查是否可以使用上述选择性提取/打开方法打开.zip文件中的所有文件?

我还能如何选择性地解压缩/打开zipfile?还有更多的pythonic方法吗?

How else can I perform selective extract/open of zipfiles? Is there a more pythonic method?

推荐答案

您的代码存在 错误.它正在打开并读取也在 inzipfile.namelist()中的文件夹名称.您可以通过以下方式简单地看到它:

There is something wrong with your code. It's opening and reading the folder names which are also in inzipfile.namelist(). You can see this by simply:

print inzipfile.namelist()

哪个会输出:

['foobar/bar/', 'foobar/bar/b.txt', 'foobar/bar/c.txt', 'foobar/foo/', 
 'foobar/foo/a.txt', 'foobar/foo/b.txt', 'foobar/']

查看它的另一种方法是使用 inzipfile.printdir()，这将导致打印以下几行内容:

Another way to see it is withinzipfile.printdir()which should result in something along the following lines being printed:

File Name                                             Modified             Size
foobar/bar/                                    2014-01-12 08:53:36            0
foobar/bar/b.txt                               2014-01-12 08:54:08           60
foobar/bar/c.txt                               2014-01-12 08:54:28           60
foobar/foo/                                    2014-01-12 08:53:02            0
foobar/foo/a.txt                               2014-01-12 08:55:04           60
foobar/foo/b.txt                               2014-01-12 08:55:24           60
foobar/                                        2014-01-12 08:52:32            0

请注意，在两种情况下，所有文件夹条目的名称均以/字符结尾.您可以使用它作为检测它们的简单方法:

Notice that in both cases the name of all folder entries end with a/character. You can use that as a simple way to detect them:

import os
import zipfile

with zipfile.ZipFile('foobar.zip', 'r') as inzipfile:
    for infile in (name for name in inzipfile.namelist() if name[-1] != '/'):
        if 'foo' in os.path.split(infile)[0]:
            print inzipfile.open(infile,'r').read(),

与之类似:

wanted = {'a.txt', 'b.txt'}  # use a set, it's faster for testing membership
import zipfile
with zipfile.ZipFile('foobar.zip','r') as inzipfile:
    for infile in (name for name in inzipfile.namelist() if name[-1] != '/'):
        if os.path.split(infile)[1] in wanted:
          print inzipfile.open(infile,'r').read()

我想想想检查档案文件的所有[file]成员是否都可以打开的唯一方法是实际尝试对每个文件进行打开:

The only way I can think of to check if all the [file] members of an archive can be opened, is to actually try doing it to each one:

def check_files(zipfilename):
    """ Check and see if all members of a .zip archive can be opened.
        Beware of vacuous truth - all members of an empty archive can be opened
    """
    def can_open(archive, membername):
        try:
            archive.open(membername, 'r')  # return value ignored
        except (RuntimeError, zipfile.BadZipfile, zipfile.LargeZipFile):
            return False
        return True

    with zipfile.ZipFile(zipfilename, 'r') as archive:
        return all(can_open(archive, membername)
                    for membername in (
                        name for name in archive.namelist() if name[-1] != '/'))

这篇关于在python中选择性提取和打开zipfile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！