How do you get the directory name from the error object on IOError in python 2.3.4?

Question

When trying to open a directory, an exception is thrown:

>>> try:
...   open( '/' )
... except IOError, e:
...   print e.filename
None

Is it possible (preferrably in python 2.3.4) for the handler of the exception to get the name of the directory from the exception object?

Why are you using such an old version of python? – Nathan Jones Jan 02 '12 at 16:01 — Nathan Jones, Jan 02 '12 at 16:01
@nathan I ask this question nearly every day... – William Pursell Jan 02 '12 at 16:07 — William Pursell, Jan 02 '12 at 16:07

philofinfinitejest · Accepted Answer · 2012-01-02T17:50:03.257

1

This is a bug in older versions of Python: http://bugs.python.org/issue4764

Supplying the open function with a mode, open('/', 'r'), might allow you to retrieve the filename via e.filename.

edited Jan 02 '12 at 17:50

answered Jan 02 '12 at 17:31

1 Answers1