Reputation: 3445

javascript regular expression to check for IP addresses

I have several ip addresses like:

115.42.150.37
115.42.150.38
115.42.150.50

What type of regular expression should I write if I want to search for the all the 3 ip addresses? Eg, if I do 115.42.150.* (I will be able to search for all 3 ip addresses)

What I can do now is something like: /[0-9]{1-3}\.[0-9]{1-3}\.[0-9]{1-3}\.[0-9]{1-3}/ but it can't seems to work well.

Thanks.

Upvotes: 89

Answers (28)

Coralin Pableo

Reputation: 1

const num=198;

const pattern =(num) => /^\[1-9][0-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5]$/.test(num) ;

console.log(pattern(num));

const x=120,r=123,f=245,h=197;

if (pattern(String(x)) && pattern(String(r)) && pattern(String(f)) && pattern(String(h))) {
  console.log(`${x}.${r}.${f}.${h}`);
    } else {
     console.log("One onset is wrong")
    }

Upvotes: 0

daemon.devin

Reputation: 138

For those trying to validate IPv4 and IPv6, the following snippet makes use of @Spudley's concise answer..

const isValidIPv4 = (ip) => {
    const blocks = ip.split('.');
    if (blocks.length !== 4) {
        return false;
    }
    for (const block of blocks) {
        const numericBlock = parseInt(block, 10);
        if (!(numericBlock >= 0 && numericBlock <= 255)) {
            return false; 
        }
    }
    return true;
}
const isValidIPv6 = (ip) => {
    const blocks = ip.split(':');
    if (blocks.length < 8) {
        return false;
    }
    for (const block of blocks) {
        if (!/^[0-9A-Fa-f]{1,4}$/.test(block)) {
            return false; 
        }
    }
    return true;
}
const isValidIP = (ip) => {
    return isValidIPv4(ip) || isValidIPv6(ip);
}

Thank you @Spudley

Upvotes: 0

Bhargav Jodhani

Reputation: 1

this is how i used ip address validation for may code, and it is working properly.

if (!ipaddress.match(/^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/)) {
 alert("Invalid Ip Address");
 return;
}

Upvotes: 0

Alexis Osmany

Reputation: 26

A regular expression for validate well conformed ip address

/^(25[0-4]|2[0-4]\d|1(\d{2})|[1-9]\d|[1-9])(\.(25[0-4]|2[0-4]\d|1(\d{2})|[1-9]\d|\d)){3}$/

Example:

const ipArr = ['\\\\115.42.150.37\\', '\\\\192.168.0.1\\', '   \\\\66.88.3', '\\\\110.234.52.124\\', '\\\\Z:\\'];

const getIpHost = (ip) => {
  if (ip) {
    const hostPortArr = ip.replace(/\\/g, '').split(':');
    return (hostPortArr[0] || '').trim();
  }
  return '';
}

const validateIp = (ip) => {
  if (!ip?.trim().length) {
    return false;
  }

  const ipAdr = getIpHost(ip);
  const regExp = /^(25[0-4]|2[0-4]\d|1(\d{2})|[1-9]\d|[1-9])(\.(25[0-4]|2[0-4]\d|1(\d{2})|[1-9]\d|\d)){3}$/;
  return regExp.test(ipAdr);
}

ipArr.forEach((ip) => {
  const isValid = validateIp(ip);
  console.log(`${ip}:`, isValid)
})

Console Output:

\\115.42.150.37\: true
\\192.168.0.1\: true
   \\66.88.3: false
\\110.234.52.124\: true
\\Z:\: false

Upvotes: 0

Ala Eddine JEBALI

Reputation: 7891

A short RegEx: ^(?:(?:^|\.)(?:2(?:5[0-5]|[0-4]\d)|1?\d?\d)){4}$

Example

const isValidIp = value => (/^(?:(?:^|\.)(?:2(?:5[0-5]|[0-4]\d)|1?\d?\d)){4}$/.test(value));


// valid
console.log("isValidIp('0.0.0.0') ? ", isValidIp('0.0.0.0'));
console.log("isValidIp('115.42.150.37') ? ", isValidIp('115.42.150.37'));
console.log("isValidIp('192.168.0.1') ? ", isValidIp('192.168.0.1'));
console.log("isValidIp('110.234.52.124') ? ", isValidIp('110.234.52.124'));
console.log("isValidIp('115.42.150.37') ? ", isValidIp('115.42.150.37'));
console.log("isValidIp('115.42.150.38') ? ", isValidIp('115.42.150.38'));
console.log("isValidIp('115.42.150.50') ? ", isValidIp('115.42.150.50'));
console.log("isValidIp('192.168.1.70') ? ", isValidIp('192.168.1.70'));

// Invalid
console.log("isValidIp('210.110') ? ", isValidIp('210.110'));
console.log("isValidIp('255') ? ", isValidIp('255'));
console.log("isValidIp('y.y.y.y') ? ", isValidIp('y.y.y.y'));
console.log("isValidIp('255.0.0.y') ? ", isValidIp('255.0.0.y'));
console.log("isValidIp('666.10.10.20') ? ", isValidIp('666.10.10.20'));
console.log("isValidIp('4444.11.11.11') ? ", isValidIp('4444.11.11.11'));
console.log("isValidIp('33.3333.33.3') ? ", isValidIp('33.3333.33.3'));
console.log("isValidIp('0192.168.1.70') ? ", isValidIp('0192.168.1.70'));

Upvotes: 8

mikemaccana

Reputation: 123610

Don't write your own regex or copy paste! You probably won't cover all edge cases (IPv6, but also octal IPs, etc). Use the is-ip package from npm:

var isIp = require('is-ip');

isIp('192.168.0.1');

isIp('1:2:3:4:5:6:7:8');

Will return a Boolean.

Upvotes: 22

Luis Aldana

Reputation: 1

well I try this, I considered cases and how the entries had to be:

function isValidIP(str) {
   let cong= /^(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])$/  
   return cong.test(str);}

Upvotes: 0

Filas Siuma

Reputation: 321

This should work:

function isValidIP(str) {
  const arr = str.split(".").filter((el) => {
    return !/^0.|\D/g.test(el);
  });

  return arr.filter((el) => el.length && el >= 0 && el <= 255).length === 4;
}

Upvotes: 1

WillianSteiangel

Reputation: 13

In addition to a solution without regex:

const checkValidIpv4 = (entry) => {
  const mainPipeline = [
    block => !isNaN(parseInt(block, 10)),
    block => parseInt(block,10) >= 0,
    block => parseInt(block,10) <= 255,
    block => String(block).length === 1
      || String(block).length > 1
      && String(block)[0] !== '0',
  ];

  const blocks = entry.split(".");
  if(blocks.length === 4
    && !blocks.every(block => parseInt(block, 10) === 0)) {
    return blocks.every(block =>
      mainPipeline.every(ckeck => ckeck(block) )
    );
  }

  return false;
}

console.log(checkValidIpv4('0.0.0.0')); //false
console.log(checkValidIpv4('0.0.0.1')); //true
console.log(checkValidIpv4('0.01.001.0')); //false
console.log(checkValidIpv4('8.0.8.0')); //true

Upvotes: 1

Ali Zedan

Reputation: 283

The answers over allow leading zeros in Ip address, and that it is not correct. For example ("123.045.067.089"should return false).

The correct way to do it like that.

 function isValidIP(ipaddress) {  
if (/^(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)\.(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)\.(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)\.(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)$/.test(ipaddress)) {  
  return (true)  
}   
return (false) }

This function will not allow zero to lead IP addresses.

Upvotes: 1

Carlos Agelvis

Reputation: 31

This is what I did and it's fast and works perfectly:

function isIPv4Address(inputString) {
    let regex = new RegExp(/^(([0-9]{1,3}\.){3}[0-9]{1,3})$/);
    if(regex.test(inputString)){
        let arInput = inputString.split(".")
        for(let i of arInput){
            if(i.length > 1 && i.charAt(0) === '0')
                return false;
            else{
                if(parseInt(i) < 0 || parseInt(i) >=256)
                   return false;
            }
        }
    }
    else
        return false;
    return true;
}

Explanation: First, with the regex check that the IP format is correct. Although, the regex won't check any value ranges.

I mean, if you can use Javascript to manage regex, why not use it?. So, instead of using a crazy regex, use Regex only for checking that the format is fine and then check that each value in the octet is in the correct value range (0 to 255). Hope this helps anybody else. Peace.

Upvotes: 3

Mac Ignacio

Reputation: 965

Simple Method

const invalidIp = ipAddress
   .split(".")
   .map(ip => Number(ip) >= 0 && Number(ip) <= 255)
   .includes(false);

if(invalidIp){
 // IP address is invalid
 // throw error here
}

Upvotes: 4

Mahdi Pedram

Reputation: 749

Below Solution doesn't accept Padding Zeros

Here is the cleanest way to validate an IP Address, Let's break it down:

Fact: a valid IP Address is has 4 octets, each octets can be a number between 0 - 255

Breakdown of Regex that matches any value between 0 - 255

25[0-5] matches 250 - 255
2[0-4][0-9] matches 200 - 249
1[0-9][0-9] matches 100 - 199
[1-9][0-9]? matches 1 - 99
0 matches 0

const octet = '(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?|0)';

Notes: When using new RegExp you should use \\. instead of \. since string will get escaped twice.

function isValidIP(str) {
  const octet = '(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?|0)';
  const regex = new RegExp(`^${octet}\\.${octet}\\.${octet}\\.${octet}$`);
  return regex.test(str);
}

Upvotes: 16

Spudley

Reputation: 168843

The regex you've got already has several problems:

Firstly, it contains dots. In regex, a dot means "match any character", where you need to match just an actual dot. For this, you need to escape it, so put a back-slash in front of the dots.

Secondly, but you're matching any three digits in each section. This means you'll match any number between 0 and 999, which obviously contains a lot of invalid IP address numbers.

This can be solved by making the number matching more complex; there are other answers on this site which explain how to do that, but frankly it's not worth the effort -- in my opinion, you'd be much better off splitting the string by the dots, and then just validating the four blocks as numeric integer ranges -- ie:

if(block >= 0 && block <= 255) {....}

Hope that helps.

Upvotes: 34

shawfire

Reputation: 79

A less stringent when testing the type not the validity. For example when sorting columns use this check to see which sort to use.

export const isIpAddress = (ipAddress) => 
    /^((\d){1,3}\.){3}(\d){1,3}$/.test(ipAddress)

When checking for validity use this test. An even more stringent test checking that the IP 8-bit numbers are in the range 0-255:

export const isValidIpAddress = (ipAddress) => 
    /^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/.test(ipAddress)

Upvotes: -1

SamWhan

Reputation: 8332

Throwing in a late contribution:

^(?!\.)((^|\.)([1-9]?\d|1\d\d|2(5[0-5]|[0-4]\d))){4}$

Of the answers I checked, they're either longer or incomplete in their verification. Longer, in my experience, means harder to overlook and therefore more prone to be erroneous. And I like to avoid repeating similar patters, for the same reason.

The main part is, of course, the test for a number - 0 to 255, but also making sure it doesn't allow initial zeroes (except for when it's a single one):

[1-9]?\d|1\d\d|2(5[0-5]|[0-4]\d)

Three alternations - one for sub 100: [1-9]?\d, one for 100-199: 1\d\d and finally 200-255: 2(5[0-5]|[0-4]\d).

This is preceded by a test for start of line or a dot ., and this whole expression is tested for 4 times by the appended {4}.

This complete test for four byte representations is started by testing for start of line followed by a negative look ahead to avoid addresses starting with a .: ^(?!\.), and ended with a test for end of line ($).

See some samples here at regex101.

Upvotes: 4

Dave Joyce

Reputation: 1

Always looking for variations, seemed to be a repetitive task so how about using forEach!

function checkIP(ip) {
  //assume IP is valid to start, once false is found, always false
  var test = true;

  //uses forEach method to test each block of IPv4 address
  ip.split('.').forEach(validateIP4);

  if (!test) 
    alert("Invalid IP4 format\n"+ip) 
  else 
    alert("IP4 format correct\n"+ip);

  function validateIP4(num, index, arr) {
    //returns NaN if not an Int
    item = parseInt(num, 10);
    //test validates Int, 0-255 range and 4 bytes of address
    // && test; at end required because this function called for each block
    test = !isNaN(item) && !isNaN(num) && item >=0 && item < 256 && arr.length==4 && test;
  }
}

Upvotes: 0

Mohsen Kadoura

Reputation: 81

If you wrtie the proper code you need only this very simple regular expression: /\d{1,3}/

function isIP(ip) {
    let arrIp = ip.split(".");
    if (arrIp.length !== 4) return "Invalid IP";
    let re = /\d{1,3}/;
    for (let oct of arrIp) {
        if (oct.match(re) === null) return "Invalid IP"
        if (Number(oct) < 0 || Number(oct) > 255)
            return "Invalid IP";
}
    return "Valid IP";
}

But actually you get even simpler code by not using any regular expression at all:

function isIp(ip) {
    var arrIp = ip.split(".");
    if (arrIp.length !== 4) return "Invalid IP";
    for (let oct of arrIp) {
        if ( isNaN(oct) || Number(oct) < 0 || Number(oct) > 255)
            return "Invalid IP";
}
    return "Valid IP";
}

Upvotes: 3

ErickBest

Reputation: 4692

May be late but, someone could try:

Example of VALID IP address

115.42.150.37
192.168.0.1
110.234.52.124

Example of INVALID IP address

210.110 – must have 4 octets
255 – must have 4 octets
y.y.y.y – only digits are allowed
255.0.0.y – only digits are allowed
666.10.10.20 – octet number must be between [0-255]
4444.11.11.11 – octet number must be between [0-255]
33.3333.33.3 – octet number must be between [0-255]

JavaScript code to validate an IP address

function ValidateIPaddress(ipaddress) {  
  if (/^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/.test(ipaddress)) {  
    return (true)  
  }  
  alert("You have entered an invalid IP address!")  
  return (false)  
}

Upvotes: 149

Andrew K

Reputation: 716

If you are using nodejs try:

require('net').isIP('10.0.0.1')

doc net.isIP()

Upvotes: 51

Tom

Reputation: 5121

/^(?!.*\.$)((?!0\d)(1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$/

Full credit to oriadam. I would have commented below his/her answer to suggest the double zero change I made, but I do not have enough reputation here yet...

change:

-(?!0) Because IPv4 addresses starting with zeros ('0.248.42.223') are valid (but not usable)
+(?!0\d) Because IPv4 addresses with leading zeros ('63.14.209.00' and '011.012.013.014') can sometimes be interpreted as octal

Upvotes: 4

oriadam

Reputation: 8629

Try this one, it's a shorter version:

^(?!0)(?!.*\.$)((1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$

Explained:

^ start of string
  (?!0)         Assume IP cannot start with 0
  (?!.*\.$)     Make sure string does not end with a dot
  (
    (
    1?\d?\d|   A single digit, two digits, or 100-199
    25[0-5]|   The numbers 250-255
    2[0-4]\d   The numbers 200-249
    )
  \.|$ the number must be followed by either a dot or end-of-string - to match the last number
  ){4}         Expect exactly four of these
$ end of string

Unit test for a browser's console:

var rx=/^(?!0)(?!.*\.$)((1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$/;
var valid=['1.2.3.4','11.11.11.11','123.123.123.123','255.250.249.0','1.12.123.255','127.0.0.1','1.0.0.0'];
var invalid=['0.1.1.1','01.1.1.1','012.1.1.1','1.2.3.4.','1.2.3\n4','1.2.3.4\n','259.0.0.1','123.','1.2.3.4.5','.1.2.3.4','1,2,3,4','1.2.333.4','1.299.3.4'];
valid.forEach(function(s){if (!rx.test(s))console.log('bad valid: '+s);});
invalid.forEach(function(s){if (rx.test(s)) console.log('bad invalid: '+s);});

Upvotes: 65

Ben McCormick

Reputation: 25728

If you want something more readable than regex for ipv4 in modern browsers you can go with

function checkIsIPV4(entry) {
  var blocks = entry.split(".");
  if(blocks.length === 4) {
    return blocks.every(function(block) {
      return parseInt(block,10) >=0 && parseInt(block,10) <= 255;
    });
  }
  return false;
}

Upvotes: 13

user826822

Reputation:

it is maybe better:

function checkIP(ip) {
    var x = ip.split("."), x1, x2, x3, x4;

    if (x.length == 4) {
        x1 = parseInt(x[0], 10);
        x2 = parseInt(x[1], 10);
        x3 = parseInt(x[2], 10);
        x4 = parseInt(x[3], 10);

        if (isNaN(x1) || isNaN(x2) || isNaN(x3) || isNaN(x4)) {
            return false;
        }

        if ((x1 >= 0 && x1 <= 255) && (x2 >= 0 && x2 <= 255) && (x3 >= 0 && x3 <= 255) && (x4 >= 0 && x4 <= 255)) {
            return true;
        }
    }
    return false;
}

Upvotes: 1

waleed.arshad

Reputation: 31

Regular expression for the IP address format:

/^(\d\d?)|(1\d\d)|(0\d\d)|(2[0-4]\d)|(2[0-5])\.(\d\d?)|(1\d\d)|(0\d\d)|(2[0-4]\d)|(2[0-5])\.(\d\d?)|(1\d\d)|(0\d\d)|(2[0-4]\d)|(2[0-5])$/;

Upvotes: 3

macf00bar

Reputation: 700

\b(?:[0-9]{1,3}\.){3}[0-9]{1,3}\b

matches 0.0.0.0 through 999.999.999.999 use if you know the seachdata does not contain invalid IP addresses

\b(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b

use to match IP numbers with accurracy - each of the 4 numbers is stored into it's own capturing group, so you can access them later

Upvotes: 1

Teja Kantamneni

Reputation: 17482

Try this one.. Source from here.

"\b(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b"

Upvotes: 16

Mikel

Reputation: 25656

And instead of

{1-3}

you should put

{1,3}

Upvotes: 2

javascript regular expression to check for IP addresses

Answers (28)

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