Question

How can I unzip a .zip file with Python into some directory output_dir and fetch a list of all the directories made by the unzipping as a result? For example, if I have:

unzip('myzip.zip', 'outdir')

outdir is a directory that might have other files/directories in it. When I unzip myzip.zip into it, I’d like unzip to return all the directories made in outdir/ as a result of the zipping. Here is my code so far:

import zipfile
def unzip(zip_file, outdir):
    """
    Unzip a given 'zip_file' into the output directory 'outdir'.
    """
    zf = zipfile.ZipFile(zip_file, "r")
    zf.extractall(outdir)

How can I make unzip return the dirs it creates in outdir? thanks.

Edit: the solution that makes most sense to me is to get ONLY the top-level directories in the zip file and then recursively walk through them which will guarantee that I get all the files made by the zip. Is this possible? The system specific behavior of namelist makes it virtually impossible to rely on

Answer 1

You can read the contents of the zip file with the namelist() method. Directories will have a trailing path separator:

>>> import zipfile
>>> zip = zipfile.ZipFile('test.zip')
>>> zip.namelist()
['dir2/', 'file1']

You can do this before or after extracting contents.

Depending on your operating environment, the result of namelist() may be limited to the top-level paths of the zip archive (e.g. Python on Linux) or may cover the full contents of the archive (e.g. IronPython on Windows).

The namelist() returns a complete listing of the zip archive contents, with directories marked with a trailing path separator. For instance, a zip archive of the following file structure:

./file1
./dir2
./dir2/dir21
./dir3
./dir3/file3
./dir3/dir31
./dir3/dir31/file31

results in the following list being returned by zipfile.ZipFile.namelist():

[ 'file1', 
  'dir2/', 
  'dir2/dir21/', 
  'dir3/', 
  'dir3/file3', 
  'dir3/dir31/', 
  'dir3/dir31/file31' ]