ip v4的正则表达式

Question

I am doing an ip v4 validator program with javascript. The ip has to be between the range of 1.0.0.0 to 255.255.255.255 My problem is that I also get the 0.0.0.0 and should not take it.

I leave you my regular expression:

Var Expression1 = new RegExp ("[0-9] | [1-9] [0-9] | 1 [0-9] {2} | 2 [0-4] [0-9] | 25 [0-5]) [3] ([0-9] | [1-9] [0-9] | 1 [0-9] {2} | 2 [0-4] [0-9] ] | 25 [0-5]) $ ");

Thank you!

Answer 1

Add "not 0.0.0.0" negative lookbehind ( ^(?!0\\.0\\.0\\.0) ) at the beginning of the line:

^(?!0\.0\.0\.0)(([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.){3}([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])$

Demo: https://regex101.com/r/dIkHRJ/2

PS

Please notice your regex is incomplete and broken. Use the corrected one from the sample above.

Answer 2

Why not use a function, it will be clearer not using a regex:

 function validateIP(ip) { var p = ip.split('.'); // split it by '.' if(p.length !== 4) return false; // if there is not for parts => false for(var i = 0; i < 4; i++) // for each part if(isNaN(p[i]) || i < 0 || i > 255) return false; // if it's not a number between 0 - 255 => false return p[3] > 0; // return true if the last part is not 0, false otherwise } alert(validateIP(prompt("IP: "))); 

Answer 3

The more accurate solution using RegExp.prototype.test() , String.prototype.split() and Array.prototype.every() functions:

 var re = /^\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}$/, validateIPv4 = function(ip) { return re.test(ip) && ip.split('.').every(function(d, idx){ var octet = Number(d); if (idx === 0) { return octet > 0; } return 0 <= octet && octet <= 255; }); }; console.log(validateIPv4('0.0.0.0')); console.log(validateIPv4('1.0.0.0')); console.log(validateIPv4('127.0.0.1')); console.log(validateIPv4('255.255.255.255'));