i want to validate the value is valid IP Address or not..!
I Used to validate like
ValidIpAddressRegex = "^(([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])$";
it's working fine, but when i give the values like 12345678
, its also return true..
How to solve this?
\d
which is more compact than [0-9]
, but it is more efficient for the regex engine if you use [0-9]
. See https://stackoverflow.com/questions/1662173 - Marathon55 2017-12-26 16:05
There is a simpler way. You just need to split the string on .
and check that every number is between 0 and 255.
Additionally, you can check for hexa and split on :
for IPv6.
Just because I think it's funny:
^(\d|[1-9]\d|1\d\d|2([0-4]\d|5[0-5]))\.(\d|[1-9]\d|1\d\d|2([0-4]\d|5[0-5]))\.(\d|[1-9]\d|1\d\d|2([0-4]\d|5[0-5]))\.(\d|[1-9]\d|1\d\d|2([0-4]\d|5[0-5]))$
Here is a regex that should handle IPs (v4).
(?<!\S)((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\b|\.\b){7}(?!\S)
Srichakradhar 2013-12-25 18:09
Looking for one for IPv4, I ended up just creating it myself. (This only handles the common dotted variant, i.e. 0.0.0.0 - 255.255.255.255)
^ # START OF STRING
(?=\d+\.\d+\.\d+\.\d+$) # Lookahead, require this format: number.number.number.number END OF STRING
(?: # Start non-capture group (number 0-255 + optional dot)
(?: # Start non-capture group (number 0-255)
25[0-5] # 250-255
| # OR
2[0-4][0-9] # 200-249
| # OR
1[0-9]{2} # 100-199
| # OR
[1-9][0-9] # 10-99
| # OR
[0-9] # 0-9
) # End non-capture group
\.? # Optional dot (enforced in correct positions by lookahead)
){4} # End non-capture group (number + optional dot), repeat 4 times
$ # END OF STRING
Without comments:
^(?=\d+\.\d+\.\d+\.\d+$)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9][0-9]|[0-9])\.?){4}$
Some code to test it:
function isValidIpv4Addr(ip) {
return /^(?=\d+\.\d+\.\d+\.\d+$)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9][0-9]|[0-9])\.?){4}$/.test(ip);
}
var testAddr = ['192.68.35.35','0.0.0.0','255.0.0.0','192.168.1.0','192.168.0.1','255.255.255.0','1.1.1.1','255.255.255.255','249.249.249.249','200.200.200.200','199.199.199.199','100.100.100.100','99.99.99.99','0.0.0.0','9.9.9.9','10.10.10.10','99.99.99.99','100.100.100.100','109.109.109.109','110.110.110.110','199.199.199.199','200.200.200.200','249.249.249.249','250.250.250.250','255.255.255.255','256.256.256.260','192.168.0.0/24','192.168..1','192.168.1','1','1.','1.1','1.1.','1.1.1','1.1.1.','1.1.1.1.','1.1.1.1.1','.1.1.1.1','01.01.01.01','09.09.09.09','1.0.0.1.0','010.1.1.1','123456','123123123123','.127.0.0.1'];
for (var i = 0; i < testAddr.length; i++) {
document.getElementById('ipv4tests').innerHTML += '<li>' + testAddr[i] + ' ' + (isValidIpv4Addr(testAddr[i]) ? '<font color="green">VALID!</font>' : '<font color="red">INVALID!</font>') + '</li>';
}
<ul id="ipv4tests"></ul>
256.256.256.259
- https://regex101.com/r/vX2hK4/1 - Nick Grealy 2015-02-17 06:12
^(?=\d+\.\d+\.\d+\.\d+(\/\d+)?$)(?:(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9][0-9]|[0-9])\.?){4}(?:\/(?:[0-9]|1[0-9]|2[0-9]|3[0-2]))?$
Nathan 2016-03-25 16:22
This works properly for all possible cases.
^(([1-9]?\d|1\d\d|2[0-5][0-5]|2[0-4]\d)\.){3}([1-9]?\d|1\d\d|2[0-5][0-5]|2[0-4]\d)$
^(([1-9]?\d|1\d\d|25[0-5]|2[0-4]\d)\.){3}([1-9]?\d|1\d\d|25[0-5]|2[0-4]\d)$
Great answer, btw - Srichakradhar 2013-12-25 05:45
I know this is old, but try this one:
/^(?:(?:2[0-4]\d|25[0-5]|1\d{2}|[1-9]?\d)\.){3}(?:2[0-4]\d|25[0-5]|1\d{2}|[1-9]?\d)(?:\:(?:\d|[1-9]\d{1,3}|[1-5]\d{4}|6[0-4]\d{3}|65[0-4]\d{2}|655[0-2]\d|6553[0-5]))?$/
I made it today for a function in php.
It handles ip's from 0.0.0.0 to 255.255.255.255 and ports from 0 to 65535.
Examples:
validates:
0.0.0.0:0
255.0.0.0
192.168.1.0:8080
does not validate:
192.168.0.0/24
192.168..1
192.168.1
I know this is a frankenregex, but still, it works!
If port doesn't matter, use this one:
/^(?:(?:2[0-4]\d|25[0-5]|1\d{2}|[1-9]?\d)\.){3}(?:2[0-4]\d|25[0-5]|1\d{2}|[1-9]?\d)$/
Try this shortened one:
^(([1-9]?\d|1\d\d|2[0-4]\d|25[0-5])(\.(?!$)|(?=$))){4}$
Here is the test case for this regex:
function verifyIp(ip)
{
return /^(([1-9]?\d|1\d\d|2[0-4]\d|25[0-5])(\.(?!$)|(?=$))){4}$/.test(ip||"");
}
["192.68.35.35","0.0.0.0","255.0.0.0","192.168.1.0","192.168.0.1","255.255.255.0","1.1.1.1","255.255.255.255","249.249.249.249","200.200.200.200","199.199.199.199","100.100.100.100","99.99.99.99","0.0.0.0","9.9.9.9","10.10.10.10","99.99.99.99","100.100.100.100","109.109.109.109","110.110.110.110","199.199.199.199","200.200.200.200","249.249.249.249","250.250.250.250","255.255.255.255","256.256.256.260","192.168.0.0/24","192.168..1","192.168.1","1","1.","1.1","1.1.","1.1.1","1.1.1.","1.1.1.1.","1.1.1.1.1",".1.1.1.1","01.01.01.01","09.09.09.09","1.0.0.1.0","010.1.1.1","123456","123123123123",".127.0.0.1"].forEach(function(item){
is_valid = verifyIp(item);
$('<div>'+item+' <span class="'+(is_valid?'correct':'wrong')+'">'+(is_valid?'VALID':'INVALID')+'</span></div>').appendTo('#result');
});
.item {
font-weight: bold;
}
.wrong {
color: red;
}
.correct {
color: green;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="result"></div>
Here is solution:
^(([01]?[0-9]?[0-9]|2([0-4][0-9]|5[0-5]))\.){3}([01]?[0-9]?[0-9]|2([0-4][0-9]|5[0-5]))$
010.1.1.1
- https://regex101.com/r/vX2hK4/1 - Nick Grealy 2015-02-17 06:14
You might also try this:
^((?:(?:^|\.)(?:\d|[1-9]\d|1\d{2}|2[0-4]\d|25[0-5])){4})$
We want the pattern to repeat exactly four times - in this case our pattern is a number in the range of 0 - 255 preceded by either a period .
or the start of the string! Since the start of the string can occur only once, the other three occurrences must be periods.
Please see this Regex 101 demo for a full explanation.
.1.1.1.1
scentos 2015-02-16 12:06
(?!^\.)
: https://regex101.com/r/vX2hK4/1 - David Faber 2015-02-16 18:38
Just extending on @DavidFaber 's excellent solution. To match an IPv4 "Dotted decimal" notation (no range/ports):
^(((1?[1-9]?|10|2[0-4])\d|25[0-5])($|\.(?!$))){4}$
Match examples: https://regex101.com/r/vX2hK4/15
Code golf anyone?
.127.0.0.1
scentos 2015-02-16 12:02
.
prefixed addresses - Nick Grealy 2015-02-16 12:50
This reg ex works well but trust me its an overkill.
To have conditional comparisons like here less then 255 its best to have combination of RegEx and conditionals.
^(([0-1]?[0-9]?[0-9]{1})|(2?[0-4]?[0-9]{1})|(25[0-5]))\.(([0-1]?[0-9]?[0-9]{1})|(2?[0-4]?[0-9]{1})|(25[0-5]))\.(([0-1]?[0-9]?[0-9]{1})|(2?[0-4]?[0-9]{1})|(25[0-5]))\.(([0-1]?[0-9]?[0-9]{1})|(2?[0-4]?[0-9]{1})|(25[0-5]))$
you can simply use this regex to validate any ip address without port number, like this format (192.168.1.1)
/^(?:(?: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]?)$/
TRY THIS,
ValidIpAddressRegex = "^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$"
trying to shorten Grealy's version
^((1?\d?\d|2[0-4]\d|25[0-5])($|\.(?!$))){4}$
note: as the previous version this doesn't correctly handle octal numbers, like 0177.0.0.1
You can also check with my given expressions as well, I had checked and written a program in java to validate the ipv4 address. It returns true if Ipv4 address is correct or vice-versa.
String pattern="^([01]?\d\d?|2[0-4]\d|25[0-5])\.([01]?\d\d?|2[0-4]\d|25[0-5])\.([01]?\d\d?|2[0-4]\d|25[0-5])\.([01]?\d\d?|2[0-4]\d|25[0-5])$"
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.Scanner;
class Solution{
public static void main(String []args){
Scanner in = new Scanner(System.in);
while(in.hasNext()){
String IP = in.next();
System.out.println(IP.matches(new MyRegex().pattern));
}
}
}
class MyRegex{
String pattern="^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.([01]?\\dCongrats, you solved this challenge!\\d?|2[0-4]\\d|25[0-5])$";
}
Colin Hebert pointed out the best solution. But no one have "explained" by supplying the code for it, so here goes ( "Just because I think it's funny:" ;)
var aIP = [
'192.168.0.1',
'255.255.255.255',
'1.2.34.647',
'256.0.0.0',
'255,0,0,0',
'123.123.123',
'1.2.3.4.5'
];
aIP.forEach(function(ipAddr) {
var a = ipAddr.split('.'),
cnt = 4;
document.write('Testing ' + ipAddr + '<br/>');
try {
a.forEach(function(v) {
if( v<0 || v>255 )
throw false;
cnt--;
});
if( cnt!=0 )
throw false;
cnt--;
document.write('- Pass!<br/>');
}
catch (e) {
document.write('- Fail!<br/>');
}
});
Try this as well:
(((?<![\d])([0-9][\.])|(?<![\d])([1-9][0-9][\.])|(?<![\d])(1[0-9]{2}[\.])|(?<![\d])(2[0-5][0-5][\.]))(([0-9][\.])|([1-9][0-9][\.])|(1[0-9]{2}[\.])|(2[0-5][0-5][\.])){2}(([0-9])(?![\d])|([1-9][0-9])(?![\d])|(1[0-9]{2})(?![\d])|(2[0-5][0-5])(?![\d])))
although this is a 5 years old question so I doubt you are still looking for the answer.
posted more info on another thread: Validating IPv4 addresses with regexp
for the pattern ex 192.168.23.28/255.255.255.0
^(25[0-5]|2[0-4][0-9]|[01]?[1-9][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[0-4]|2[0-4][0-9]|[01]?[1-9][0-9]?)\/(((128|192|224|240|248|252|254)\.0\.0\.0)|(255\.(0|128|192|224|240|248|252|254)\.0\.0)|(255\.255\.(0|128|192|224|240|248|252|254)\.0)|(255\.255\.255\.(0|128|192|224|240|248|252|254)))$
for the pattern ex 192.168.26.82/24 or 192.168.23.28/255.255.255.0
^(25[0-5]|2[0-4][0-9]|[01]?[1-9][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[0-4]|2[0-4][0-9]|[01]?[1-9][0-9]?)\/([1-9]|1[0-9]|2[0-9]|3[0-2]|(((128|192|224|240|248|252|254)\.0\.0\.0)|(255\.(0|128|192|224|240|248|252|254)\.0\.0)|(255\.255\.(0|128|192|224|240|248|252|254)\.0)|(255\.255\.255\.(0|128|192|224|240|248|252|254))))$
for the pattern ex 192.168.26.28
^(25[0-5]|2[0-4][0-9]|[01]?[1-9][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[0-4]|2[0-4][0-9]|[01]?[1-9][0-9]?)$
for the netmask ex 255.255.255.0
^(((128|192|224|240|248|252|254)\.0\.0\.0)|(255\.(0|128|192|224|240|248|252|254)\.0\.0)|(255\.255\.(0|128|192|224|240|248|252|254)\.0)|(255\.255\.255\.(0|128|192|224|240|248|252|254)))$
The OP asked for validation of an IP address. The phrasing of the Q almost certainly implies IPv4 (as opposed to IPv6). I've already commented on the OP about how far that validity checking might go, as well as applauding one responder for taking an non-RE approach. As I also wanted to (under some circumstances) test for validity for the address if I use it on a public server, I came up with the following JS:
function isSimpleIPv4( ip, u=true ) {
if ((ip === undefined) || (ip === null) || (ip.length > 15)) return false;
var p = ip.split(\'.\');
if (p.length != 4) return false;
p.forEach( function(v,k){p[k]=Number(v);} );
if (isNaN(p[0]) || isNaN(p[1]) || isNaN(p[2]) || isNaN(p[3]) ) return false;
if ((p[0] < 1) || (p[0] > 255) || (p[1] < 0) || (p[1] > 255) || (p[2] < 0) || (p[2] > 255) || (p[3] < 0) || (p[3] > 255)) return false;
if (!u) return true;
if ((p[0] > 223)) return 'multicast';
if ((p[0] == 127)) return 'loopback';
if ((p[0] == 10)) return 'RFC1918';
if ((p[0] == 192) && (p[1] == 168)) return 'RFC1918';
if ((p[0] == 172) && (p[1] >= 16) && (p[1] <= 31)) return 'RFC1918';
return true;
}
If one wants to check for "useful" addresses, then only the string IP is needed, otherwise, if just checking for 0-255.0-255.0-255.0-255, then call the function with the IP-string and false. In the former case, the function will return true/false/ when is the "failing" cause of the usefulness. RFC1918 addresses intended by a website visitor won't be reachable from a public server and may even compromise one of one's own servers (as might addresses in the loopback range 127.x.x.x). Equally, using a multicast address is not helpful either. If you're interested in usefulness checking but don't care about the cause, then you'll just need to do if (isSimpleIPv4(ipString) !== true) console.log('Not a valid and useful IP address');
IPv4 has 4 blocks of numbers from 0 to 255 that could contain padding zeroes on the left. Each block is spaced with a dot.
Short and simple IPv4 validator:
var block "([0-1]{0,1}[0-9]{1,2}|2[0-4][0-9]|25[0-5]|)";
var ipv4 = "(" + block +"\\.){3}" + block ;
Validates any IP Like:
import { Component } from '@angular/core';
import { FormBuilder, FormGroup, Validators } from '@angular/forms';
@Component({
selector: 'app-root',
templateUrl: './app.component.html',
styleUrls: ['./app.component.scss'],
})
export class AppComponent {
ipranging="";
testForm1: FormGroup;
testForm2: FormGroup;
constructor(private fb: FormBuilder){
}
ngOnInit(): void {
const ipPattern =
"(?:(?: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]?)";
this.testForm1 = this.fb.group({
inp: ['', Validators.pattern(ipPattern)],
inp3: ['', Validators.pattern(ipPattern)]
});
this.testForm2 = this.fb.group({
inp: ['', Validators.pattern(ipPattern)],
inp2: ['', Validators.pattern(ipPattern)],
inp3: ['', Validators.pattern(ipPattern)]
});
this.testForm2.setValidators(this.comparisionValidator);
}
public comparisionValidator(group: FormGroup) : any{
const control1 = group.controls['inp'];
const control2 = group.controls['inp2'];
var control1array = control1.value.split('.');
var control2array = control2.value.split('.');
if(parseInt(control1array[0]) > parseInt(control2array[0]) ){
group.controls['inp3'].setErrors({ 'value2GreaterThanValue1': true });
console.log(group);
}
else if(parseInt(control1array[1]) > parseInt(control2array[1]) ){
group.controls['inp3'].setErrors({ 'value2GreaterThanValue1': true });
console.log(group);
}
else if(parseInt(control1array[2]) > parseInt(control2array[2]) ){
group.controls['inp3'].setErrors({ 'value2GreaterThanValue1': true });
console.log(group);
}
else if(parseInt(control1array[3]) > parseInt(control2array[3]) ){
group.controls['inp3'].setErrors({ 'value2GreaterThanValue1': true });
console.log(group);
}
else {
group.controls['inp3'].setErrors({ 'value2GreaterThanValue1': false });
console.log(group);
}
}
}
<div style="text-align:left">
<h2>Choose if you want to enter a single ip or range of ip's</h2>
<select [(ngModel)]="ipranging">
<option selected disabled value="none"> -- select an option -- </option>
<option value='ip'>Single Ip address</option>
<option value="range">Ip range</option>
</select>
</div>
<form *ngIf="ipranging === 'ip'" novalidate [formGroup]="testForm1" class="render">
<label>IP Address:
<input formControlName="inp" placeholder='0.0.0.0'/></label>
<input formControlName="inp3" hidden/>
<!-- <p *ngIf="testForm.controls.inp.status == 'INVALID' && testForm.controls.inp.value != ''" >Invalid</p>
<p *ngIf="testForm.controls.inp2.status == 'INVALID' && testForm.controls.inp2.value != ''" >Invalid</p> -->
<p *ngIf="testForm1.controls.inp.value != '' && testForm1.controls.inp.status == 'INVALID'" >Invalid</p>
</form>
<form *ngIf="ipranging === 'range'" novalidate [formGroup]="testForm2" class="render">
<label>Starting IP:
<input formControlName="inp" placeholder='0.0.0.0'/></label>
<label>
Ending IP:
<input formControlName="inp2" placeholder='0.0.0.0'/></label>
<input formControlName="inp3" hidden/>
<!-- <p *ngIf="testForm.controls.inp.status == 'INVALID' && testForm.controls.inp.value != ''" >Invalid</p>
<p *ngIf="testForm.controls.inp2.status == 'INVALID' && testForm.controls.inp2.value != ''" >Invalid</p> -->
<p *ngIf="testForm2.controls.inp.value != '' && testForm2.controls.inp.status == 'INVALID' || testForm2.controls.inp2.value != '' && testForm2.controls.inp2.status == 'INVALID'" >Invalid</p>
<p *ngIf="testForm2.controls.inp3.errors.value2GreaterThanValue1 == true">Starting IP is larger than the ending IP</p>
</form>
This should work
^(\\d{1,3})\\.(\\d{1,3})\\.(\\d{1,3})\\.(\\d{1,3})$