javascript regular expression to check for IP addresses

May be late but, someone could try:

Example of VALID IP address

115.42.150.37192.168.0.1110.234.52.124

Example of INVALID IP address

210.110 – must have 4 octets255 – must have 4 octetsy.y.y.y – only digits are allowed255.0.0.y – only digits are allowed666.10.10.20 – 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)  }  

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);});

If you are using nodejs try:

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

doc net.isIP()