I have a PC behind a firewall, and I'm trying to programmatically
determine the IP address visible from outside the firewall.
If I do this:
>>> import socket
>>> socket.gethostbyname(socket.gethostname())
'127.0.0.1'
>>> socket.gethostbyname_ex(socket.gethostname())
('localhost.localdomain', ['localhost'], ['127.0.0.1'])
I get my internal IP address, which is not what I want.
Other tricks, like parsing the output of os.system('/sbin/ifconfig eth0')
also give me my internal IP address.
I found this post on comp.lang.python a few years ago:
http://mail.python.org/pipermail/python-list/2003-March/192495.html
which seems like it _should_ do what I want, but it doesn't: I get an
exception.
>>> from httplib import HTTPConnection
>>> from xml.dom.ext.reader.Sax import FromXmlStream
>>> conn = HTTPConnection('xml.showmyip.com')
>>> conn.request('GET', '/')
>>> doc = FromXmlStream(conn.getresponse())
>>> print doc.getElementsByTagName('ip')[0].firstChild.data
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "/usr/lib/python2.4/UserList.py", line 28, in __getitem__
def __getitem__(self, i): return self.data[i]
IndexError: list index out of range
>>> conn.close()
I don't know how to fix it so that it works.
Can anyone help me fix that code snippet, or suggest another (better) way
to get the externally visible IP address?
--
Steven D'Aprano
--
http://mail.python.org/mailman/listinfo/python-list
