Regular expression that matches valid IPv6 addresses

Answer 1

Using Ruby? Try this:

/^(((?=.*(::))(?!.*\3.+\3))\3?|[\dA-F]{1,4}:)([\dA-F]{1,4}(\3|:\b)|\2){5}(([\dA-F]{1,4}(\3|:\b|$)|\2){2}|(((2[0-4]|1\d|[1-9])?\d|25[0-5])\.?\b){4})\z/i

Answer 2

For PHP 5.2+ users filter_var works great.

I know this doesn't answer the original question (specifically a regex solution), but I post this in the hope it may help someone else in the future.

$is_ip4address = (filter_var($ip, FILTER_VALIDATE_IP, FILTER_FLAG_IPV4) !== FALSE);
$is_ip6address = (filter_var($ip, FILTER_VALIDATE_IP, FILTER_FLAG_IPV6) !== FALSE);

Answer 3

If you use Perl try Net::IPv6Addr

use Net::IPv6Addr;

if( defined Net::IPv6Addr::is_ipv6($ip_address) ){
  print "Looks like an ipv6 address\n";
}

NetAddr::IP

use NetAddr::IP;

my $obj = NetAddr::IP->new6($ip_address);

Validate::IP

use Validate::IP qw'is_ipv6';

if( is_ipv6($ip_address) ){
  print "Looks like an ipv6 address\n";
}

Answer 4

In Java, you can use the library class sun.net.util.IPAddressUtil:

IPAddressUtil.isIPv6LiteralAddress(iPaddress);

Answer 5

Just matching local ones from an origin with square brackets included. I know it's not as comprehensive but in javascript the other ones had difficult to trace issues primarily that of not working, so this seems to get me what I needed for now. extra capitals A-F aren't needed either.

^\[([0-9a-fA-F]{1,4})(\:{1,2})([0-9a-fA-F]{1,4})(\:{1,2})([0-9a-fA-F]{1,4})(\:{1,2})([0-9a-fA-F]{1,4})(\:{1,2})([0-9a-fA-F]{1,4})\]

Jinnko's version is simplified and better I see.

Answer 6

The regex allows the use of leading zeros in the IPv4 parts.

Some Unix and Mac distros convert those segments into octals.

I suggest using 25[0-5]|2[0-4]\d|1\d\d|[1-9]?\d as an IPv4 segment.