CODEWITHSUNDEEP
javascriptvalidationextjsrange
user2569803
user2569803

Reputation: 637

JavaScript IP Range Validation is second ip is greater than first ip

Hello I am a javascript noob and was writing a validator for an IP range. For example 1.1.1.1-2.2.2.2 is a valid range however I want to make sure that the first IP is not greater than the second IP. 2.2.2.2-1.1.1.1 should return false. What is the best strategy to go about this problem, should I try to convert each IP string to a map of integers and then compare them one by one? Should I iterate a for loop 1-4 and compare the first value before the . and then the second etc. 

ipRangeValidator: function(value)
{
    if (!value)
        return true;
    var split = value.split('-');
    if (!split || split.length > 2 || split.length === 0) {
        return 'Input only an IP or single IP Range';
    }

    var ips = [];
    for (var x = 0; x < split.length; ++x) {
        var ip = split[x];
        if (isValidIpV6Address(ip)) { 
        } else if (isValidIpV4Address(ip)) {
        } else {
            return ip + ' is not a valid IP address';
        }
        ips.push(ip);
    }
    if (ips.length > 1) {
       if(true) {            //The Line I am struggling with
           return 'The first IP is greater than the second IP';
       }
    }
    return true;
}

Upvotes: 0

Views: 967

Answers (2)

Yevhen Horbunkov
Yevhen Horbunkov

Reputation: 15530

Would something, like that work for you?

const validRange = '1.1.1.1-2.2.2.2',
      invalidRange = '2.2.2.2-1.1.1.1',
      anotherInvalidRange = '1.1.3.1-1.1.2.1',
      invalidIp = '1.1.300.1-1.1.1.1',
      
      isRangeValid = range => {
        const [rangeStart, rangeEnd] = range.split('-'),
              [s1,s2,s3,s4] = rangeStart.split('.'),
              [e1,e2,e3,e4] = rangeEnd.split('.')
        return [s1,s2,s3,s4].some(o => o>255 || o<0) || [e1,e2,e3,e4].some(o => o>255 || o<0) ?
          'range contains invalid IP' :
          s1>e1 || s2>e2 || s3>e3 || s4>e4 ?
          'range start is greater than range end' :
          'range is valid'
      }
      
console.log(isRangeValid(validRange))
console.log(isRangeValid(invalidRange))
console.log(isRangeValid(anotherInvalidRange))
console.log(isRangeValid(invalidIp))
.as-console-wrapper{min-height:100%;}

Upvotes: 4

mplungjan
mplungjan

Reputation: 178026

Does this help?

const pad = (num) => String("00"+num).slice(-3);
const ipRange = rng => {
  let [ip1,ip2] = rng.split("-");
  ip1 = ip1.split(".").map(num => pad(num)).join('.')
  ip2 = ip2.split(".").map(num => pad(num)).join('.')
  return ip1>ip2;
};
console.log(
  ipRange("1.1.1.1-2.2.2.2")
)  
console.log(
  ipRange("2.2.2.2-1.1.1.1")
)

Upvotes: 0

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