Matching an IP to a CIDR mask in PHP 5?

If only using IPv4:

• use ip2long() to convert the IPs and the subnet range into long integers
• convert the /xx into a subnet mask
• do a bitwise 'and' (i.e. ip & mask)' and check that that 'result = subnet'

something like this should work:

function cidr_match($ip, $range)
{
    list ($subnet, $bits) = explode('/', $range);
    if ($bits === null) {
        $bits = 32;
    }
    $ip = ip2long($ip);
    $subnet = ip2long($subnet);
    $mask = -1 << (32 - $bits);
    $subnet &= $mask; # nb: in case the supplied subnet wasn't correctly aligned
    return ($ip & $mask) == $subnet;
}

In a similar situation, I ended up using symfony/http-foundation.

When using this package, your code would look like:

$ips = array('10.2.1.100', '10.2.1.101', '10.5.1.100', '1.2.3.4');

foreach($ips as $IP) {
    if (\Symfony\Component\HttpFoundation\IpUtils::checkIp($IP, '10.2.0.0/16')) {
        print "you're in the 10.2 subnet\n";
    }
}

It also handles IPv6.

Link: https://packagist.org/packages/symfony/http-foundation

I found many of these methods breaking after PHP 5.2. However the following solution works on versions 5.2 and above:

function cidr_match($ip, $cidr)
{
    list($subnet, $mask) = explode('/', $cidr);

    if ((ip2long($ip) & ~((1 << (32 - $mask)) - 1) ) == ip2long($subnet))
    { 
        return true;
    }

    return false;
}

Example results

cidr_match("1.2.3.4", "0.0.0.0/0"):         true
cidr_match("127.0.0.1", "127.0.0.1/32"):    true
cidr_match("127.0.0.1", "127.0.0.2/32"):    false

Source http://www.php.net/manual/en/function.ip2long.php#82397.