JavaScript: Is IP In One Of These Subnets?

Question

So I have ~12600 subnets:

eg. 123.123.208.0/20

and an IP.

I can use a SQLite Database or an array or whatever

There was a similar question asked about a month ago, however I am not looking for checking one IP against one subnet but a bunch of subnets (obviously the most efficient way, hopefully not O(total subnets)) :)

How can I check that the IP is one of in one of these subnets, I need true or false not the subnet if that helps optimisation.

There are similar subnets in the current list eg.: (actual extract)

123.123.48.0/22 <-- not a typo
123.123.48.0/24 <-- not a typo
123.123.90.0/24
123.123.91.0/24
123.123.217.0/24

In total they range from 4.x.y.z to 222.x.y.z

Answer 1

The best approach is IMO making use of bitwise operators. For example, 123.123.48.0/22 represents (123<<24)+(123<<16)+(48<<8)+0 (=2071670784; this might be a negative number) as a 32 bit numeric IP address, and -1<<(32-22) = -1024 as a mask. With this, and likewise, your test IP address converted to a number, you can do:

(inputIP & testMask) == testIP

For example, 123.123.49.123 is in that range, as 2071671163 & -1024 is 2071670784

So, here are some tool functions:

function IPnumber(IPaddress) {
    var ip = IPaddress.match(/^(\d+)\.(\d+)\.(\d+)\.(\d+)$/);
    if(ip) {
        return (+ip[1]<<24) + (+ip[2]<<16) + (+ip[3]<<8) + (+ip[4]);
    }
    // else ... ?
    return null;
}

function IPmask(maskSize) {
    return -1<<(32-maskSize)
}

test:

(IPnumber('123.123.49.123') & IPmask('22')) == IPnumber('123.123.48.0')

yields true.

In case your mask is in the format '255.255.252.0', then you can use the IPnumber function for the mask, too.

Answer 2

Try this:

var ip2long = function(ip){
    var components;

    if(components = ip.match(/^(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3})$/))
    {
        var iplong = 0;
        var power  = 1;
        for(var i=4; i>=1; i-=1)
        {
            iplong += power * parseInt(components[i]);
            power  *= 256;
        }
        return iplong;
    }
    else return -1;
};

var inSubNet = function(ip, subnet)
{   
    var mask, base_ip, long_ip = ip2long(ip);
    if( (mask = subnet.match(/^(.*?)\/(\d{1,2})$/)) && ((base_ip=ip2long(mask[1])) >= 0) )
    {
        var freedom = Math.pow(2, 32 - parseInt(mask[2]));
        return (long_ip > base_ip) && (long_ip < base_ip + freedom - 1);
    }
    else return false;
};

Usage:

inSubNet('192.30.252.63', '192.30.252.0/22') => true
inSubNet('192.31.252.63', '192.30.252.0/22') => false

Answer 3

Convert the lower ip and the upper ip in the range to integers and store the range in the db then make sure both columns are indexed.

Off the top of my head (pseudo code):

function ipmap(w,x,y,z) {
  return 16777216*w + 65536*x + 256*y + z;
}

var masks = array[ipmap(128,0,0,0), ipmap(196,0,0,0), ..., ipmap(255,255,255,255)]

function lowrange(w, x, y, z, rangelength) {
  return ipmap(w, x, y, z) & masks[rangelength]
}

function hirange(w, x, y, z, rangelength) {
  return lowrange(w, x, y, z, ,rangelength) + ipmap(255,255,255,255) - masks[rangelength];
}

That ought to do it.

To find whether a particular ip falls in any of the ranges, convert it to an integer and do:

SELECT COUNT(*) FROM ipranges WHERE lowrange <= 1234567 AND 1234567 <= highrange

The query optimizer should be able to speed this up greatly.

Answer 4

I managed to solve this by using the node netmask module. You can check if an IP belongs to a subnet by making something like this:

import { Netmask } from 'netmask'

const block = new Netmask('123.123.208.0/20')
const ip = '123.123.208.0'
console.log(block.contains(ip))

Will here print true.

You can install it by using:

npm i --save netmask

Answer 5

Functions IPnumber and IPmask are nice, however I would rather test like:

(IPnumber('123.123.49.123') & IPmask('22')) == (IPnumber('123.123.48.0')  & IPmask('22'))

Because for each address, you only need to take into account the network part of the address. Hence doing IPmask('22') will zero-out the computer part of the address and you should do the same with the network address.