If I have an IP address like 2001:4860:4860::8888
How can I get the fully qualified domain in the format foo.ip6.arpa ?
EDIT: Both the solutions so far give me google-public-dns-a.google.com - maybe Reverse DNS was the wrong name. For this I'd expect the output to be something like 8.8.8.8.0...etc.ip6.arpa
IPy provides methods for what you want:
>>> from IPy import IP >>> ip = IP('127.0.0.1') >>> ip.reverseName() '1.0.0.127.in-addr.arpa.' Works for both IPv4 and IPv6, although the original IPy has a few bugs for IPv6. I created a fork with some extensions and fixes at https://github.com/steffann/python-ipy which you can use as long as the fixes haven't been merged back into the original code.
You can of course also use the getnameinfo function in built-in socket module:
>>> import socket >>> socket.getnameinfo(('2001:4860:4860::8888', 0), 0) ('google-public-dns-a.google.com', '0') >>> socket.getnameinfo(('127.0.0.1', 0), 0) ('localhost', '0') You need to provide a host+port tuple, but you can provide 0 for the port, and you'll get the hostname back.
socket.getnameinfo is not really reverse DNS resolution (which would mean to ask only/exactly the DNS). With socket.getnameinfo other sources like /etc/hosts may be taken into account as well (and typically are even before asking the DNS). And DNS may not even be asked at all.using dnspython.
from dns import resolver,reversename addr=reversename.from_address("2001:4860:4860::8888") str(resolver.query(addr,"PTR")[0]) 
.ip6.arpaformat?