How can I parse a host:port pair in Python

Question

Suppose I have a string of the of the format host:port, where :port is optional. How can I reliably extract the two components?

The host can be any of:

• A hostname (localhost, www.google.com)
• An IPv4 literal (1.2.3.4)
• An IPv6 literal ([aaaa:bbbb::cccc]).

In other words, this is the standard format used across the internet (such as in URIs: complete grammar at https://www.rfc-editor.org/rfc/rfc3986#section-3.2, excluding the "User Information" component).

So, some possible inputs, and desired outputs:

'localhost' -> ('localhost', None)
'my-example.com:1234' -> ('my-example.com', 1234)
'1.2.3.4' -> ('1.2.3.4', None)
'[0abc:1def::1234]' -> ('[0abc:1def::1234]', None)

Answer 1

Well, this is Python, with batteries included. You have mention that the format is the standard one used in URIs, so how about urllib.parse?

import urllib.parse

def parse_hostport(hp):
    # urlparse() and urlsplit() insists on absolute URLs starting with "//"
    result = urllib.parse.urlsplit('//' + hp)
    return result.hostname, result.port

This should handle any valid host:port you can throw at it.

Answer 2

Came up with a dead simple regexp that seems to work in most cases:

def get_host_pair(value):
    return re.search(r'^(.*?)(?::(\d+))?$', value).groups()

get_host_pair('localhost')
get_host_pair('localhost:80')
get_host_pair('[::1]')
get_host_pair('[::1]:8080')

It probably doesn't work when the base input is invalid however