Unzipping directory structure with python

Question

I have a zip file which contains the following directory structure:

dir1\\dir2\\dir3a
dir1\\dir2\\dir3b

I\'m trying to unzip it and maintai

Answer 1

I tried this out, and can reproduce it. The extractall method, as suggested by other answers, does not solve the problem. This seems like a bug in the zipfile module to me (perhaps Windows-only?), unless I'm misunderstanding how zipfiles are structured.

testa\
testa\testb\
testa\testb\test.log
> test.zip

>>> from zipfile import ZipFile
>>> zipTest = ZipFile("C:\\...\\test.zip")
>>> zipTest.extractall("C:\\...\\")
Traceback (most recent call last):
  File "", line 1, in 
  File "...\zipfile.py", line 940, in extractall
  File "...\zipfile.py", line 928, in extract
  File "...\zipfile.py", line 965, in _extract_member
IOError: [Errno 2] No such file or directory: 'C:\\...\\testa\\testb\\test.log'

If I do a printdir(), I get this (first column):

>>> zipTest.printdir()
File Name
testa/testb/
testa/testb/test.log

If I try to extract just the first entry, like this:

>>> zipTest.extract("testa/testb/")
'C:\\...\\testa\\testb'

On disk, this results in the creation of a folder testa, with a file testb inside. This is apparently the reason why the subsequent attempt to extract test.log fails; testa\testb is a file, not a folder.

Edit #1: If you extract just the file, then it works:

>>> zipTest.extract("testa/testb/test.log")
'C:\\...\\testa\\testb\\test.log'

Edit #2: Jeff's code is the way to go; iterate through namelist; if it's a directory, create the directory. Otherwise, extract the file.