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# vim: tabstop=4 shiftwidth=4 softtabstop=4
# Copyright 2012 OpenStack Foundation.
# All Rights Reserved.
#
# Licensed under the Apache License, Version 2.0 (the "License"); you may
# not use this file except in compliance with the License. You may obtain
# a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS, WITHOUT
# WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
# License for the specific language governing permissions and limitations
# under the License.
"""
Network-related utilities and helper functions.
"""
import urlparse
def parse_host_port(address, default_port=None):
"""Interpret a string as a host:port pair.
An IPv6 address MUST be escaped if accompanied by a port,
because otherwise ambiguity ensues: 2001:db8:85a3::8a2e:370:7334
means both [2001:db8:85a3::8a2e:370:7334] and
[2001:db8:85a3::8a2e:370]:7334.
>>> parse_host_port('server01:80')
('server01', 80)
>>> parse_host_port('server01')
('server01', None)
>>> parse_host_port('server01', default_port=1234)
('server01', 1234)
>>> parse_host_port('[::1]:80')
('::1', 80)
>>> parse_host_port('[::1]')
('::1', None)
>>> parse_host_port('[::1]', default_port=1234)
('::1', 1234)
>>> parse_host_port('2001:db8:85a3::8a2e:370:7334', default_port=1234)
('2001:db8:85a3::8a2e:370:7334', 1234)
"""
if address[0] == '[':
# Escaped ipv6
_host, _port = address[1:].split(']')
host = _host
if ':' in _port:
port = _port.split(':')[1]
else:
port = default_port
else:
if address.count(':') == 1:
host, port = address.split(':')
else:
# 0 means ipv4, >1 means ipv6.
# We prohibit unescaped ipv6 addresses with port.
host = address
port = default_port
return (host, None if port is None else int(port))
def urlsplit(url, scheme='', allow_fragments=True):
"""Parse a URL using urlparse.urlsplit(), splitting query and fragments.
This function papers over Python issue9374 when needed.
The parameters are the same as urlparse.urlsplit.
"""
scheme, netloc, path, query, fragment = urlparse.urlsplit(
url, scheme, allow_fragments)
if allow_fragments and '#' in path:
path, fragment = path.split('#', 1)
if '?' in path:
path, query = path.split('?', 1)
return urlparse.SplitResult(scheme, netloc, path, query, fragment)
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