Sorting IPv4 and IPv6 address in a Database

Question

I have a field for storing IP addresses and I'm using this code to convert an IPv4 address to an integer so I can save a sortable version of the address in a separate bigint field to achieve more natural sorting (based on this ):

octets = ip.split('.')
return (int(octets[0]) * 256**3) + (int(octets[1]) * 256**2) + (int(octets[2]) * 256) + (int(octets[3]))

How can I do something similar with IPv6 addresses which is too big for a bigint field?

And is there a simple way to do the integer conversion for IPv6? Python3 seems to provide this with the "ipaddress" module, but I'm using 2.7. I want to support the various ways of leaving zero's out of the address.

Update : I'm using Django 1.5

Answer 1

I ended up using this code to convert the IPs to integers:

import struct, socket

try:
    return struct.unpack('!I', socket.inet_pton(socket.AF_INET, ip))[0]
except socket.error:
    try:
        hi, lo = struct.unpack('!QQ', socket.inet_pton(socket.AF_INET6, ip))
        return (hi << 64) | lo
    except socket.error:
        return 0

Using a DecimalField worked, as recommended by kroolik

Answer 2

I believe I'd split IPv4 addresses on ., and sort as a list. Similarly for IPv6, I'd try splitting on :, and sorting them as a list too.

And wrapping this up as a class would probably be a good idea, in case the behavior needs to change someday. That way, if it does change, the changes aren't scattered across your code.