Question

I’m trying to unzip files in Python. I’m using the following function:

def unzip(source_filename, dest_dir):
    zf = zipfile.ZipFile(source_filename)
    for member in zf.infolist():
        # Path traversal defense copied from
        # http://hg.python.org/cpython/file/tip/Lib/http/server.py#l789
        words = filter(None, member.filename.split('/'))
        path = dest_dir
        for word in words[:-1]:
            drive, word = os.path.splitdrive(word)
            head, word = os.path.split(word)
            if word in (os.curdir, os.pardir): continue
            path = os.path.join(path, word)
        zf.extract(member, path)

When unzipping a .zip with a directory called “gui”, I get the following error:

Traceback (most recent call last):
  File "MCManager.py", line 137, in add
    unzip(addedFilepath, dirUnzipped)
  File "MCManager.py", line 19, in unzip
    zf.extract(member, path)
  File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/zipfile.py", line 928, in extract
  File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/zipfile.py", line 962, in _extract_member
  File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/os.py", line 157, in makedirs
OSError: [Errno 20] Not a directory: '/Users/student/Library/Application Support/minecraft/temp/unzipped/gui/gui'

Is this a problem with ZipFile()?

Answer 1

Checking the documentation for the method extract, you will find that the path variable that is received by your method through the variable dest_dir must be the path to a dir which you want the contents of the zip files you iterate to be extracted to.

As you have stated on you question, gui is a file, not a directory. And to be completely honest, you statement

When unzipping a .zip with a file called “gui (…)”

doesn’t make very much sense.

Additionally, you may want to take a look at the os.mkdir method, so that you can create directories on demand.