Handling Windows-specific exceptions in platform-independent way

Question

Consider the following Python exception:

  [...]
    f.extractall()
  File "C:\Python26\lib\zipfile.py", line 935, in extractall
    self.extract(zipinfo, path, pwd)
  File "C:\Python26\lib\zipfile.py", line 923, in extract
    return self._extract_member(member, path, pwd)
  File "C:\Python26\lib\zipfile.py", line 957, in _extract_member
    os.makedirs(upperdirs)
  File "C:\Python26\lib\os.py", line 157, in makedirs
    mkdir(name, mode)
WindowsError: [Error 267] The directory name is invalid: 'C:\\HOME\\as\
\pypm-infinitude\\scratch\\b\\slut-0.9.0.zip.work\\slut-0.9\\aux'

I want to handle this particular exception - i.e., WindowsError with error number 267. However, I cannot simply do the following:

try:
    do()
except WindowsError, e:
    ...

Because that would not work on Unix systems where WindowsError is not even defined in the exceptions module.

Is there an elegant way to handle this error?

Answer 1

Here's my current solution, but I slightly despise using non-trivial code in a except block:

        try:
            f.extractall()
        except OSError, e:
            # http://bugs.python.org/issue6609
            if sys.platform.startswith('win'):
                if isinstance(e, WindowsError) and e.winerror == 267:
                    raise InvalidFile, ('uses Windows special name (%s)' % e)
            raise

Answer 2

If you need to catch an exception with a name that might not always exist, then create it:

if not getattr(__builtins__, "WindowsError", None):
    class WindowsError(OSError): pass

try:
    do()
except WindowsError, e:
    print "error"

If you're on Windows, you'll use the real WindowsError class and catch the exception. If you're not, you'll create a WindowsError class that will never be raised, so the except clause doesn't cause any errors, and the except clause will never be invoked.