How To Simulate Zipfile.open In Python 2.5?

I want to extract a file from a zip to a specific path, ignoring the file path in the archive. This is very easy in Python 2.6 (my docstring is longer than the code) import shutil

Solution 1:

Haven't tested this bit, but I use something extremely similar in Python 2.4

import zipfile

def extract_from_zip(name, dest_path, zip_file):
    dest_file = open(dest_path, 'wb')
    dest_file.write(zip_file.read(name))
    dest_file.close()

extract_from_zip('path/to/file/in/archive.dat', 
        'output.txt', 
        zipfile.ZipFile('test.zip', 'r'))

Solution 2:

I know I am a bit late to the party for this question, but was having the exact same problem.

The solution I used was to copy the python 2.6.6 version of zipfile and put in a folder (I called it python_fix) and import that instead:

python_fix/zipfile.py

Then in code:

import python_fix.zipfileas zipfile

From there I was able to use the 2.6.6 version of zipfile with the python 2.5.1 interpreter (the 2.7.X versions fail on the "with" with this version")

Hope this helps someone else using ancient technology.

Solution 3:

Given my constraints, it looks like the answer was given in my question: parse the ZipFile structure yourself and use zlib.decompressobj to unzip the bytes once you've found them.

If you don't have (/suffer from) my constraints, you can find better answers here:

1. If you can, just upgrade Python 2.5 to 2.6 (or later!), as suggested in a comment by Daenyth.
2. If you only have small files in the zip which can be 100% loaded in memory, use ChrisAdams' answer
3. If you can introduce a dependency on an external utility, make an appropriate system call to /usr/bin/unzip or similar, as suggested in Vlad's answer