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?

Asked By: Tapefreak

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Source

Answer 1

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)

Answered By: Vinko Vrsalovic

Answer 2

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

Answered By: ekhumoro

Answer 3

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()

Answered By: Ovilia

Answer 4

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()

Answered By: Sled

Answer 5

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()

Answered By: Rajkumar S

How to unzip a file with Python 2.4?

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