JavaScript Regular Expression to Check for Ip Addresses

javascript regular expression to check for IP addresses

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.

Finding an IP address inside of a javascript string

In regular expressions ^ and $ mean start and end of the line. So, your expression would only capture the IP if it is the only thing in the message.

Remove them and it will work as expected.

I recommend playing with your regular expressions o RegExr as it has a tool for explaining the payload.

Validating IPv4 addresses with regexp

You've already got a working answer but just in case you are curious what was wrong with your original approach, the answer is that you need parentheses around your alternation otherwise the (\.|$) is only required if the number is less than 200.

'\b((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)(\.|$)){4}\b'
^ ^

Regex for find All ip address except IP address starts with 172

. in a regex is character that matches everything. To use it in this context, you must escape it.

Also to limit it to just ip addresses that start with 172, simply hardcode it into your regex like so:

^172\.\d{1,3}\.\d{1,3}.\d{1,3}$

Regular expression visualization

Debuggex Demo

You could then use this to filter out any matches already made.


Alternatively, if you're not starting with a list of ip addresses, you could use a negative look-ahead to grab them all straight away.

^(?!172)\d{1,3}\.\d{1,3}\.\d{1,3}.\d{1,3}$

Regular expression visualization

Debuggex Demo


Be a little careful in that this may match more than ip addresses - for example 400.660.226.602 would be captured - even though real IP4 addresses do not contain numbers higher than 255. Perhaps this won't affect your use case - but it's something to remember.


As per the comments below, if you are searching for IP addresses anywhere in the document, rather than on their own line, use \b instead of ^ and $

\b(?!172)\d{1,3}\.\d{1,3}\.\d{1,3}.\d{1,3}\b

Regular expression visualization

Debuggex Demo

This would match log formats for example, which contain an ip address within the message rather than on it's own line.

[10:01:22] Connection from 10.14.242.211 established.

How to validate an ip address with custom format using regular expression

Here's something you could try:

function customFormat(val) {  return /^(?=\d+\.\d+\.\d+\.\d+)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9][0-9]|[0-9])\.?){4} - ((?! -).)+ - \d{1,3}$/.test(val);}var testFormat = '213.39.59.78 - Public3 address.info - 24';document.getElementById('regex').innerHTML += testFormat + ' ' + (customFormat(testFormat) ? 'Valid Format!' : 'Invalid Format!');
<ul id="regex"></ul>

javascript regex for ip address with asterisk masking

I think what you are looking for is a negative look ahead to make sure no number follows an asterisk.

Like so: (\*(?!.*\d))

working example:

var ips = [  '127.*.*.*',  '127.0.*.*',  '127.0.0.*',  '127.*.*.1',  '127.*.0.1',  '127.0.*.1'];
var regex = /^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|(\*(?!.*\d)))(\.|$)){4}$/;
for(var i = 0; i < ips.length; i++){ console.log(ips[i] + ': ' + regex.test(ips[i]));}

Extract ip address from a string using regex

You have defined r as string, initialize it as regular expression.

var r = /\b\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\b/;

var r = /\b\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\b/; //http://www.regular-expressions.info/examples.html

var a = "http://www.example.com/landing.aspx?referrer=10.11.12.13";

var t = a.match(r);
console.log(t[0])

Javascript and Regex for valid IP with Wildcards

You could use this regex from this SO answer:

var regex = /^((((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]?))|(((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]?)\.){2}\*)|(((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){1}\*))$/g;

it matches all this schemes:

1.2.*
1.2.3.*
1.2.3.4

Regex101: https://regex101.com/r/QnfVc4/1

so your code'd be like this:

function ValidateIPaddress(ipaddress) {  
if (/^((((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]?))|(((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]?)\.){2}\*)|(((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){1}\*))$/g.test(ipaddress)) {
return (true);
}
alert("You have entered an invalid IP address!")
return (false)
}

Demo:

function ValidateIPaddress(ipaddress) {    if (/^((((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]?))|(((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]?)\.){2}\*)|(((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){1}\*))$/g.test(ipaddress)) {      return (true);  }    //alert("You have entered an invalid IP address!")    return (false)  }
console.log("test for 127.0.0.1 ->", ValidateIPaddress("127.0.0.1"))console.log("test for 127.0.0.* ->", ValidateIPaddress("127.0.0.*"))console.log("test for 127.0.* ->", ValidateIPaddress("127.0.*"))console.log("test for 127.* ->", ValidateIPaddress("127.*"))console.log("test for 127.0 ->", ValidateIPaddress("127.0"))


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