javascript regular expression to check for IP addr

2020-01-24 06:06发布

问题:

I have several ip addresses like:

  1. 115.42.150.37
  2. 115.42.150.38
  3. 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.

回答1:

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.



回答2:

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)  
}  


回答3:

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);});


回答4:

If you are using nodejs try:

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

doc net.isIP()



回答5:

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"


回答6:

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;
}


回答7:

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);
}


回答8:

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

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

isIp('192.168.0.1');

Will return a Boolean.

Downvoters: care to explain why using an actively maintained library is better than copy pasting from a website?



回答9:

My version as es6 method, return true for valid IPs, false otherwise

isIP(ip) {
  if (typeof(ip) !== 'string')
    return false;
  if (!ip.match(/\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}/)) {
    return false;
  }
  return ip.split('.').filter(octect => octect >= 0 && octect <= 255).length === 4;
}


回答10:

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



回答11:

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])$/;


回答12:

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.



回答13:

And instead of

{1-3}

you should put

{1,3}


回答14:

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";
}


回答15:

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) ? true : false);


// 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'));

// 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'));



回答16:

\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



回答17:

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;
}    


回答18:

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;
  }
}


回答19:

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)