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.

Solution 1:

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)  
}

Solution 2:

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);});

Solution 3:

If you are using nodejs try:

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

doc net.isIP()

Solution 4:

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.

Solution 5:

Don't write your own regex or copy paste! You probably won't cover all edge ceses (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.

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