commit | d8aef06cc3ca8a6fd8570cda665b2bb879a29390 | [log] [tgz] |
---|---|---|
author | Philipp Hagemeister <phihag@phihag.de> | Fri Oct 18 01:30:15 2019 |
committer | Philipp Hagemeister <phihag@phihag.de> | Fri Oct 18 01:30:15 2019 |
tree | 95a9111e245a7093fdbebca8b7f8afb7438d2ea2 | |
parent | f069b1572b25d71404bca3e6ae195f44c8f6e8e9 [diff] |
release 1.0.23
Python 3.3+'s ipaddress for Python 2.6, 2.7, 3.2.
This repository tracks the latest version from cpython, e.g. ipaddress from cpython 3.8 as of writing.
Note that just like in Python 3.3+ you must use character strings and not byte strings for textual IP address representations:
>>> from __future__ import unicode_literals >>> ipaddress.ip_address('1.2.3.4') IPv4Address(u'1.2.3.4')
or
>>> ipaddress.ip_address(u'1.2.3.4') IPv4Address(u'1.2.3.4')
but not:
>>> ipaddress.ip_address(b'1.2.3.4') Traceback (most recent call last): File "<stdin>", line 1, in <module> File "ipaddress.py", line 163, in ip_address ' a unicode object?' % address) ipaddress.AddressValueError: '1.2.3.4' does not appear to be an IPv4 or IPv6 address. Did you pass in a bytes (str in Python 2) instead of a unicode object?