From Bugzilla Helper: User-Agent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.3b) Gecko/20030111 Description of problem: IPv6 addresses in URIs have to be written like http://[fe80::207:e9ff:fe9b]:8000 but passing such a string to the httplib functions produces errors like File "//usr/lib/python2.2/httplib.py", line 969, in __init__ self._setup(self._connection_class(host, port, strict)) File "//usr/lib/python2.2/httplib.py", line 491, in __init__ self._set_hostport(host, port) File "//usr/lib/python2.2/httplib.py", line 502, in _set_hostport raise InvalidURL("nonnumeric port: '%s'" % host[i+1:]) httplib.InvalidURL: nonnumeric port: ':207:e9ff:fe9b]:8000' Version-Release number of selected component (if applicable): How reproducible: Always Steps to Reproduce: 1.start python 2.run the script import xmlrpclib server = xmlrpclib.Server("http://fe80::207:e9ff:fe9b:8000") server.echo("test") It's not necessary to have a server running, you'll not get as far as using it. Actual Results: Traceback (most recent call last): File "<stdin>", line 1, in ? File "//usr/lib/python2.2/xmlrpclib.py", line 821, in __call__ return self.__send(self.__name, args) File "//usr/lib/python2.2/xmlrpclib.py", line 975, in __request verbose=self.__verbose File "//usr/lib/python2.2/xmlrpclib.py", line 833, in request h = self.make_connection(host) File "//usr/lib/python2.2/xmlrpclib.py", line 862, in make_connection return httplib.HTTP(host) File "//usr/lib/python2.2/httplib.py", line 969, in __init__ self._setup(self._connection_class(host, port, strict)) File "//usr/lib/python2.2/httplib.py", line 491, in __init__ self._set_hostport(host, port) File "//usr/lib/python2.2/httplib.py", line 502, in _set_hostport raise InvalidURL("nonnumeric port: '%s'" % host[i+1:]) httplib.InvalidURL: nonnumeric port: ':207:e9ff:fe9b]:8000' Expected Results: connection to xmlrpc server Additional info: See RFC 2732 for info on IPv6 URIs.
I'll have to check if there is any statement python 2.2 would be IPv6 compliant, anywhere. Will send upstream.
Looking at the report again, the URL is wrong. Replace the second line in the script from the receipe with server = xmlrpclib.Server("http://[fe80::207:e9ff:fe9b]:8000") Result is still the same (with python-2.2.3-7).
This specific issue seems to be fixed in 2.4.2-3.2.1 and maybe earlier.