How to unzip a file with Python 2.4?

Question

I'm having a hard time figuring out how to unzip a zip file with 2.4. extract() is not included in 2.4. I'm restricted to using 2.4.4 on my server.

Can someone please provide a simple code example?

you can use "``" backtick operator or some other way execute system function and unzip your file — , Oct 18 '11 at 11:37
If you found this question but are using a newer version of python do this: zfile = zipfile.ZipFile(file_to_extract) zfile.extractall(target_dir) — Fabian, Jun 18 '14 at 07:55
@Fabian: You could skip the zFile variable and just have `zipfile.ZipFile(file_to_extract).extractall(target_dir)` - this suffers from the same problem that your code does, though, which is that you didn't `close()` the `ZipFile` afterwards which could lead to some OS problems (IE, you won't be able to delete the file, because it'll appear as in use by Python.) — ArtOfWarfare, Jan 27 '15 at 17:07

Vinko Vrsalovic · Accepted Answer · 2014-02-11T16:28:44.123

52

You have to use namelist() and extract(). Sample considering directories

import zipfile
import os.path
import os
zfile = zipfile.ZipFile("test.zip")
for name in zfile.namelist():
  (dirname, filename) = os.path.split(name)
  print "Decompressing " + filename + " on " + dirname
  if not os.path.exists(dirname):
    os.makedirs(dirname)
  zfile.extract(name, dirname)

edited Feb 11 '14 at 16:28

answered Oct 18 '11 at 11:49

Vinko Vrsalovic

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2

Ended up copying this one and noticed one little thing. At least if you zip your file with win7 "sendTo zip file" option, and your zip file contains nested folders, you need to change os.mkdir(dirname) -> os.makedirs(dirname). Otherwise you might get exceptions(No such file or directory), as zip file contains only leaf folders – fastfox Jul 25 '13 at 12:03
1

What if `name` is a directory (not regular file)? I came across this case. – Nawaz Aug 08 '13 at 09:47
This solution is not going to work without further modifications. Two issues are : - using fd.write not in loop (so we may not write entire file if it is large). Using zfile.extract(name, concrete_dir) seems to be better - using os.mkdir will not create subdirectories. os.makedirs(concrete_dir) looks better – Dmitriusan Feb 06 '14 at 16:20
Use zfile.extract(name, '.') at last line instead to preserve zipped folder structure. – hurturk Jul 08 '14 at 20:08
@VinkoVrsalovic How do you save the files you've extracted? – NumenorForLife Apr 28 '15 at 05:17

score 12 · Answer 2 · answered Jan 30 '13 at 03:12

There's some problem with Vinko's answer (at least when I run it). I got:

IOError: [Errno 13] Permission denied: '01org-webapps-countingbeads-422c4e1/'

Here's how to solve it:

# unzip a file
def unzip(path):
    zfile = zipfile.ZipFile(path)
    for name in zfile.namelist():
        (dirname, filename) = os.path.split(name)
        if filename == '':
            # directory
            if not os.path.exists(dirname):
                os.mkdir(dirname)
        else:
            # file
            fd = open(name, 'w')
            fd.write(zfile.read(name))
            fd.close()
    zfile.close()

Shouldn't that be `fd = open(name, 'wb')` in case some of the zipped files are images or otherwise binary files? — Sled, Apr 23 '13 at 18:57

score 3 · Answer 3 · edited May 23 '17 at 10:31

Modifying Ovilia's answer so that you can specify the destination directory as well:

def unzip(zipFilePath, destDir):
    zfile = zipfile.ZipFile(zipFilePath)
    for name in zfile.namelist():
        (dirName, fileName) = os.path.split(name)
        if fileName == '':
            # directory
            newDir = destDir + '/' + dirName
            if not os.path.exists(newDir):
                os.mkdir(newDir)
        else:
            # file
            fd = open(destDir + '/' + name, 'wb')
            fd.write(zfile.read(name))
            fd.close()
    zfile.close()

score 1 · Answer 4 · answered Oct 18 '11 at 12:32

Not fully tested, but it should be okay:

import os
from zipfile import ZipFile, ZipInfo

class ZipCompat(ZipFile):
    def __init__(self, *args, **kwargs):
        ZipFile.__init__(self, *args, **kwargs)

    def extract(self, member, path=None, pwd=None):
        if not isinstance(member, ZipInfo):
            member = self.getinfo(member)
        if path is None:
            path = os.getcwd()
        return self._extract_member(member, path)

    def extractall(self, path=None, members=None, pwd=None):
        if members is None:
            members = self.namelist()
        for zipinfo in members:
            self.extract(zipinfo, path)

    def _extract_member(self, member, targetpath):
        if (targetpath[-1:] in (os.path.sep, os.path.altsep)
            and len(os.path.splitdrive(targetpath)[1]) > 1):
            targetpath = targetpath[:-1]
        if member.filename[0] == '/':
            targetpath = os.path.join(targetpath, member.filename[1:])
        else:
            targetpath = os.path.join(targetpath, member.filename)
        targetpath = os.path.normpath(targetpath)
        upperdirs = os.path.dirname(targetpath)
        if upperdirs and not os.path.exists(upperdirs):
            os.makedirs(upperdirs)
        if member.filename[-1] == '/':
            if not os.path.isdir(targetpath):
                os.mkdir(targetpath)
            return targetpath
        target = file(targetpath, "wb")
        try:
            target.write(self.read(member.filename))
        finally:
            target.close()
        return targetpath

score -1 · Answer 5 · answered Sep 23 '13 at 06:46

I am testing in Python 2.7.3rc2 and the the ZipFile.namelist() is not returning an entry with just the sub directory name for creating a sub directory, but only a list of file names with sub directory, as follows:

['20130923104558/control.json', '20130923104558/test.csv']

Thus the check

if fileName == '':

does not evaluate to True at all.

So I modified the code to check if the dirName exists inside destDir and to create dirName if it does not exist. File is extracted only if fileName part is not empty. So this should take care of the condition where a directory name can appear in ZipFile.namelist()

def unzip(zipFilePath, destDir):
    zfile = zipfile.ZipFile(zipFilePath)
    for name in zfile.namelist():
        (dirName, fileName) = os.path.split(name)
        # Check if the directory exisits
        newDir = destDir + '/' + dirName
        if not os.path.exists(newDir):
            os.mkdir(newDir)
        if not fileName == '':
            # file
            fd = open(destDir + '/' + name, 'wb')
            fd.write(zfile.read(name))
            fd.close()
    zfile.close()

This is not the answer if you are running python 2.7 see my comment on the question — Fabian, Jun 18 '14 at 07:56

How to unzip a file with Python 2.4?

5 Answers5

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