Efficient regex for Canadian postal code function - javascript

var regex = /[A-Za-z]\d[A-Za-z] ?\d[A-Za-z]\d/;
var match = regex.exec(value);
if (match){
if ( (value.indexOf("-") !== -1 || value.indexOf(" ") !== -1 ) && value.length() == 7 ) {
return true;
} else if ( (value.indexOf("-") == -1 || value.indexOf(" ") == -1 ) && value.length() == 6 ) {
return true;
}
} else {
return false;
}
The regex looks for the pattern A0A 1B1.
true tests:
A0A 1B1
A0A-1B1
A0A1B1
A0A1B1C << problem child
so I added a check for "-" or " " and then a check for length.
Is there a regex, or more efficient method?


User kind, postal code strict, most efficient format:
/^[ABCEGHJ-NPRSTVXY]\d[ABCEGHJ-NPRSTV-Z][ -]?\d[ABCEGHJ-NPRSTV-Z]\d$/i
Allows:
h2t-1b8
h2z 1b8
H2Z1B8
Disallows:
Z2T 1B8 (leading Z)
H2T 1O3 (contains O)
Leading Z,W or to contain D, F, I, O, Q or U


Add anchors to your pattern:
var regex = /^[A-Za-z]\d[A-Za-z][ -]?\d[A-Za-z]\d$/;
^ means "start of string" and $ means "end of string". Adding these anchors will prevent the C from slipping in to the match since your pattern will now expect a whole string to consist of 6 (sometimes 7--as a space) characters. This added bonus should now alleviate you of having to subsequently check the string length.
Also, since it appears that you want to allow hyphens, you can slip that into an optional character class that includes the space you were originally using. Be sure to leave the hyphen as either the very first or very last character; otherwise, you will need to escape it (using a leading backslash) to prevent the regex engine from interpreting it as part of a character range (e.g. A-Z).


This one handles us and ca codes.
function postalFilter (postalCode) {
if (! postalCode) {
return null;
}
postalCode = postalCode.toString().trim();
var us = new RegExp("^\\d{5}(-{0,1}\\d{4})?$");
var ca = new RegExp(/([ABCEGHJKLMNPRSTVXY]\d)([ABCEGHJKLMNPRSTVWXYZ]\d){2}/i);
if (us.test(postalCode.toString())) {
return postalCode;
}
if (ca.test(postalCode.toString().replace(/\W+/g, ''))) {
return postalCode;
}
return null;
}
// these 5 return null
console.log(postalFilter('1a1 a1a'));
console.log(postalFilter('F1A AiA'));
console.log(postalFilter('A12345-6789'));
console.log(postalFilter('W1a1a1')); // no "w"
console.log(postalFilter('Z1a1a1')); // ... or "z" allowed in first position!
// these return canada postal less space
console.log(postalFilter('a1a 1a1'));
console.log(postalFilter('H0H 0H0'));
// these return unaltered
console.log(postalFilter('H0H0H0'));
console.log(postalFilter('a1a1a1'));
console.log(postalFilter('12345'));
console.log(postalFilter('12345-6789'));
console.log(postalFilter('123456789'));
// strip spaces
console.log(postalFilter(' 12345 '));


You have a problem with the regex StatsCan has posted the rules for what is a valid Canadian postal code:
The postal code is a six-character code defined and maintained by
Canada Post Corporation (CPC) for the purpose of sorting and
delivering mail. The characters are arranged in the form ‘ANA NAN’,
where ‘A’ represents an alphabetic character and ‘N’ represents a
numeric character (e.g., K1A 0T6). The postal code uses 18 alphabetic
characters and 10 numeric characters. Postal codes do not include the
letters D, F, I, O, Q or U, and the first position also does not make
use of the letters W or Z.
The regex should be if you wanted it strict.
/^[ABCEGHJ-NPRSTVXY][0-9][ABCEGHJ-NPRSTV-Z] [0-9][ABCEGHJ-NPRSTV-Z][0-9]$/
Also \d means number not necessarily 0-9 there may be the one errant browser that treats it as any number in unicode space which would likely cause issues for you downstream.
from: https://trajano.net/2017/05/canadian-postal-code-validation/


This is a function that will do everything for you in one shot. Accepts AAA BBB and AAABBB with or without space.
function go_postal(){
let postal = $("#postal").val();
var regex = /^[A-Za-z]\d[A-Za-z][ -]?\d[A-Za-z]\d$/;
var pr = regex .test(postal);
if(pr === true){
//all good
} else {
// not so much
}
}


function postalFilter (postalCode, type) {
if (!postalCode) {
return null;
}
postalCode = postalCode.toString().trim();
var us = new RegExp("^\\d{5}(-{0,1}\\d{4})?$");
// var ca = new RegExp(/^((?!.*[DFIOQU])[A-VXY][0-9][A-Z])|(?!.*[DFIOQU])[A-VXY][0-9][A-Z]\ ?[0-9][A-Z][0-9]$/i);
var ca = new RegExp(/^[ABCEGHJKLMNPRSTVXY]\d[ABCEGHJKLMNPRSTVWXYZ]( )?\d[ABCEGHJKLMNPRSTVWXYZ]\d$/i);
if(type == "us"){
if (us.test(postalCode.toString())) {
console.log(postalCode);
return postalCode;
}
}
if(type == "ca")
{
if (ca.test(postalCode.toString())) {
console.log(postalCode);
return postalCode;
}
}
return null;
}


regex = new RegExp(/^[ABCEGHJ-NPRSTVXY]\d[ABCEGHJ-NPRSTV-Z][-]?\d[ABCEGHJ-NPRSTV-Z]\d$/i);
if(regex.test(value))
return true;
else
return false;
This is a shorter version of the original problem, where value is any text value. Furthermore, there is no need to test for value length.

Related

Validate number formats in a contact form (javascript)

I have a function to validate phone number in a contact form, but i need to be able to put in "xxx xxx xxxx" for example, and not just "xxxxxxxx"
The number format should be:
xxx xxx xxxx
xxx-xxx-xxxx
xxx.xxx.xxxx
function validatePhone() {
var phone = document.getElementById("phone").value;
if (phone.length == 0) {
var w = document.getElementById("phoneError").textContent;
alert(w);
return false;
}
if (phone.length != 10) {
var r = document.getElementById("phoneError").textContent;
alert(r);
return false;
}
// THIS IS NOT WORKING
if (
!phone.match(/^[0-9]{10}$/) ||
!phone.match(/^\d{3}-\d{3}-\d{4}$/) ||
!phone.match(/^\d{3}.\d{3}.\d{4}$/)
) {
var t = document.getElementById("phoneError").textContent;
alert(t);
return false;
}
}

Two things: First, you are mixing up AND and OR:
if (
!phone.match(/^[0-9]{10}$/) ||
!phone.match(/^\d{3}-\d{3}-\d{4}$/) ||
!phone.match(/^\d{3}.\d{3}.\d{4}$/)
) {
As soon as one of the conditions fails, it will return false (which is basically always). You want this if to apply, when none of the expressions matches, e.g. when all of them are false. Therefor, you have to use && instead of ||. Not a AND not b AND not c.
Second: your 3rd regex is a bit off: . means "any character", so this regex would also match "123x123y1234". You need to escape the dot with a backslash: /^\d{3}\.\d{3}\.\d{4}$/
Also, you can improve this code significantly. You have 5 conditions, which could all be handled in one (if you want to allow the input of "123.123 234", otherwise you will have to do it using 3 regex). And for just checking if a regex matches a string, you maybe should use test(), because it is just slightly faster (it won't matter in your case, but just out of principle).
You can reduce your code to:
if (/^\d{3}[\s-.]\d{3}[\s-.]\d{4}$/.test(document.getElementById("phone").value) === false) {
alert (document.getElementById("phoneError").textContent);
return false;
}

Find all instances and display alert - part 2, now with regex

Thanks for your help with my earlier question:
How to find all instances and display in alert
Now I discover that I need to include some invalid character validation.
I'm trying to figure out how to include a set of regex invalid characters as part of the validation that will also show up in the same alert/textbox/whatever as the "too long/too short" validation.
So, I have a textbox which users will type or paste comma separated values such as AAAAAAA,BBBBBBB,CCCCCCCC,DDDDDDDD
And they cannot be more or less than seven characters long and they can only include certain characters.
I currently have have two separate pieces of Javascript that I'm trying to now combine:
var Invalidchars = "1234567890!##$%^&*()+=[]\\\';./{}|\":<>?";
for (var i = 0; i < document.getElementById("TextBox1").value.length; i++) {
if (Invalidchars.indexOf(document.getElementById("TextBox").value.charAt(i)) != -1){
alert
and this
var val = document.getElementById("Textbox1").value,
err = $.grep(val.split(','), function(a) { return a.length != 7; });
if (err.length) {
alert("All entries must be seven (7) characters in length. Please correct the following entries: \n" + err);
return false;
}
return true;
Any help is much appreciated!
=================================================
SOLUTION
Took a while, but using Tenub's code (which didn't quite combine my two sets code, but was close enough), I finally figured out how to merge my two sets of code into one. Here's the code if anyone is ever interested in using it:
var val = document.getElementById("TextBox1").value,
err = $.grep(val.split(','), function(a) {return (a.length = (!/^[^0-9!##$%^&*()+=;.\/\{}|:<>\\?\[\]\'\"]{7}$/.test(a)));});
if (err.length){
document.getElementById("DIV1").style.display = "inline-block";
document.getElementById("TextBox2").value = err.join(',');
return callback (false);
}
document.getElementById("DIV1").style.display = "none";
return true;

The answer is as simple as it is elegant:
var val = document.getElementById("Textbox1").value;
if(!/[^0-9!##$%^&*()+=;./{}|:<>?\[\]\\\'\"]{7}/.test(val)) {
// handle invalid value
}
This tests that the string is 7 characters in length and does not contain any character within the brackets after the "^" (also some characters are escaped with a "\").
You can test in console:
/[^0-9!##$%^&*()+=;./{}|:<>?\[\]\\\'\"]{7}/.test('adfFDKZ'); // returns true
/[^0-9!##$%^&*()+=;./{}|:<>?\[\]\\\'\"]{7}/.test('adf(DKZ'); // returns false

Try this:
/*
* This regex matches all the invalid characters. I escaped the
* special characters.
*/
var regex = /.*[0-9!##\$%\^&\*\(\)\+=\[\]\\';\./\{\}\|":\<\>\?]+.*/;
var text = document.getElementById("TextBox1").value;
/* Test for match...much faster than a for-loop under any circumstances */
if (text.matches(regex)) {
alert("Invalid characters present. Please correct the input");
return false;
}
/* split on delimiter */
var err = $.grep(val.split(','), function(a) { return a.length != 7; });
if (err.length) {
alert("All entries must be seven (7) characters in length. Please correct the following entries: \n" + err);
return false;
}
Please tell me if there are any bugs in this. Also, the only real way to test for this in one step is to set up an enormously long regex. Also, with only one check, it would make it a little harder to guide the user to make the right correction. I will mention that.

Javascript check for spaces

I have this function but I want to check for spaces only in the front and back, not in the middle before i sent back what can i do with it...
function validateNumeric() {
var val = document.getElementById("tbNumber").value;
var validChars = '0123456789.';
for(var i = 0; i < val.length; i++){
if(validChars.indexOf(val.charAt(i)) == -1){
alert('Please enter valid number');
return false;
}
}
return true;
}

Time for regular expressions.
function startsOrEndsWithWhitespace(str)
{
return /^\s|\s$/.test(str);
}
Tests:
> /^\s|\s$/.test('123454')
false
> /^\s|\s$/.test('123 454')
false
> /^\s|\s$/.test(' 123454')
true
> /^\s|\s$/.test(' 123454 ')
true
> /^\s|\s$/.test('123454 ')
true
if i dont wanna accept 1 1 what do i have to change
function containsWhitespace(str)
{
return /\s/.test(str);
}
Tests:
> /\s/.test('123454')
false
> /\s/.test('123 454')
true
> /\s/.test(' 123454')
true
> /\s/.test('123454 ')
true
> /\s/.test(' 123454 ')
true
> /\s/.test(' 123 454 ')
true

For a really simple solution, if you can use jQuery, use jQuery.trim() and compare the trimmed string with the original. If not equal, then there were spaces so the number is invalid.

function trim (myString)
{
return myString.replace(/^\s+/g,'').replace(/\s+$/g,'')
}
source

To trim your string you can write something like this, as it's been said before:
function trim(str){
return str.replace(/^\s+|\s+$/g), '');
}
But why bother?
Want you really want is:
function validateNumeric(str) {
return !isNaN(parseFloat(str));
}
Note your original code accepts something like "..." or "7.8..9" as being numeric, which is wrong.
Update: kennebec has called my attention to the fact that parseFloat() will ignore trailing garbage at the end of string. So I call your attention to this alternative given in an answer to question "Validate numbers in JavaScript - IsNumeric()":
function isNumber(n) {
return !isNaN(parseFloat(n)) && isFinite(n);
}
(Original credit goes to CMS).

function validateNumeric() {
var val = document.getElementById("tbNumber").value;
if (!/^\s*(?:\d+(?:\.\d*)?|\.\d+)\s*$/.test(val)) {
alert('Please enter a valid number');
return false;
}
return true;
}
(?:\d+(?:\.\d*)|\.\d+) breaks down as follows:
\d+ is any number of digits, e.g. 123
(\.\d*)? optionally matches a fraction, e.g. .25 and . or blank but not .1.2
\.\d+ matches a fraction without an integer part as in .5 but not 1.5.
(?:abc|def) groups things together and matches either abc or def
/^\s*(?:\d+(?:\.\d*)|\.\d+)\s*$/ means any number of spaces followed by one or more decimal digits followed by any number of spaces. So it does what your validChars loop did plus allows spaces at the start and end.

Regular expression for DN addresses

How to write a regular expression in javascript that must follow the conditions
All segment in the DN address should follow the sequence cn=<name>,ou=<name>,o=<bic8>,o=swift
All segments should be separated with ,.
The DN address should have maximum of 100 characters.
No space is allowed.
Minimum of 2 and maximum of 10 segments are allowed in a DN address.
The <name> part must contain minimum of 2 characters and maximum 20 alphanumeric characters. The characters should be in lower case. Only one special character is allowed to be used i.e. -(Hypen).
The DN address will have maximum 2 numbers. The <name> part can contain maximum of 2 numerical digits.
Thanks in advance

I think .split() is a lot easier to use in this case.
First split the entire string on the ,'s and then split every separate segment of the resulting array on the ='s.
Especially on a well defined spec as this, split is more then enough to handle it.

Untested code follows, don't blame me if it blows up your computer:
var parseDn(str)
var m = /^cn=(.*?),ou=(.*?),o=(.*?),o=swift$/.exec(str);
if (!m) { return null; } // (a) and (b).
if (s.length > 100) { return null; } // (c).
if (/\s/.exec(s)) { return null; } // (d).
var x = {cn:m[1], ou:m[2], o:m[3]};
var isValidName = function(s) { return (/^[a-z-]{2,20}$/).exec(s); }
if (!isValidName(x.cn) || !isValidName(x.ou) || !isValidName(x.o)) {
return null; // (f).
}
var countNumbers = function(s) { return s.replace(/\D/g, "").length; }
if (countNumbers(x.cn)>2 || countNumbers(x.ou)>2 || countNumbers(x.o)>2) {
return null; // (g).
}
return x; // => {"cn":"name", "ou":"name", "o":"bic8"}
}
Note that (e) and a few of the points regarding "segments" are completely unchecked since the description is vague. But this should get you started...

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.

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)
}

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()

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.

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

Try this one.. Source from here.
"\b(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]?)\b"

Below Solution doesn't accept Padding Zeros
Here is the cleanest way to validate an IP Address, Let's break it down:
Fact: a valid IP Address is has 4 octets, each octets can be a number between 0 - 255
Breakdown of Regex that matches any value between 0 - 255
25[0-5] matches 250 - 255
2[0-4][0-9] matches 200 - 249
1[0-9][0-9] matches 100 - 199
[1-9][0-9]? matches 1 - 99
0 matches 0
const octet = '(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?|0)';
Notes: When using new RegExp you should use \\. instead of \. since string will get escaped twice.
function isValidIP(str) {
const octet = '(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?|0)';
const regex = new RegExp(`^${octet}\\.${octet}\\.${octet}\\.${octet}$`);
return regex.test(str);
}

If you want something more readable than regex for ipv4 in modern browsers you can go with
function checkIsIPV4(entry) {
var blocks = entry.split(".");
if(blocks.length === 4) {
return blocks.every(function(block) {
return parseInt(block,10) >=0 && parseInt(block,10) <= 255;
});
}
return false;
}

A short RegEx: ^(?:(?:^|\.)(?:2(?:5[0-5]|[0-4]\d)|1?\d?\d)){4}$
Example
const isValidIp = value => (/^(?:(?:^|\.)(?:2(?:5[0-5]|[0-4]\d)|1?\d?\d)){4}$/.test(value) ? true : false);
// valid
console.log("isValidIp('0.0.0.0') ? ", isValidIp('0.0.0.0'));
console.log("isValidIp('115.42.150.37') ? ", isValidIp('115.42.150.37'));
console.log("isValidIp('192.168.0.1') ? ", isValidIp('192.168.0.1'));
console.log("isValidIp('110.234.52.124' ? ", isValidIp('110.234.52.124'));
console.log("isValidIp('115.42.150.37') ? ", isValidIp('115.42.150.37'));
console.log("isValidIp('115.42.150.38') ? ", isValidIp('115.42.150.38'));
console.log("isValidIp('115.42.150.50') ? ", isValidIp('115.42.150.50'));
// Invalid
console.log("isValidIp('210.110') ? ", isValidIp('210.110'));
console.log("isValidIp('255') ? ", isValidIp('255'));
console.log("isValidIp('y.y.y.y' ? ", isValidIp('y.y.y.y'));
console.log(" isValidIp('255.0.0.y') ? ", isValidIp('255.0.0.y'));
console.log("isValidIp('666.10.10.20') ? ", isValidIp('666.10.10.20'));
console.log("isValidIp('4444.11.11.11') ? ", isValidIp('4444.11.11.11'));
console.log("isValidIp('33.3333.33.3') ? ", isValidIp('33.3333.33.3'));

/^(?!.*\.$)((?!0\d)(1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$/
Full credit to oriadam. I would have commented below his/her answer to suggest the double zero change I made, but I do not have enough reputation here yet...
change:
-(?!0) Because IPv4 addresses starting with zeros ('0.248.42.223') are valid (but not usable)
+(?!0\d) Because IPv4 addresses with leading zeros ('63.14.209.00' and '011.012.013.014') can sometimes be interpreted as octal

Simple Method
const invalidIp = ipAddress
.split(".")
.map(ip => Number(ip) >= 0 && Number(ip) <= 255)
.includes(false);
if(invalidIp){
// IP address is invalid
// throw error here
}

Regular expression for the IP address format:
/^(\d\d?)|(1\d\d)|(0\d\d)|(2[0-4]\d)|(2[0-5])\.(\d\d?)|(1\d\d)|(0\d\d)|(2[0-4]\d)|(2[0-5])\.(\d\d?)|(1\d\d)|(0\d\d)|(2[0-4]\d)|(2[0-5])$/;

If you wrtie the proper code you need only this very simple regular expression: /\d{1,3}/
function isIP(ip) {
let arrIp = ip.split(".");
if (arrIp.length !== 4) return "Invalid IP";
let re = /\d{1,3}/;
for (let oct of arrIp) {
if (oct.match(re) === null) return "Invalid IP"
if (Number(oct) < 0 || Number(oct) > 255)
return "Invalid IP";
}
return "Valid IP";
}
But actually you get even simpler code by not using any regular expression at all:
function isIp(ip) {
var arrIp = ip.split(".");
if (arrIp.length !== 4) return "Invalid IP";
for (let oct of arrIp) {
if ( isNaN(oct) || Number(oct) < 0 || Number(oct) > 255)
return "Invalid IP";
}
return "Valid IP";
}

Throwing in a late contribution:
^(?!\.)((^|\.)([1-9]?\d|1\d\d|2(5[0-5]|[0-4]\d))){4}$
Of the answers I checked, they're either longer or incomplete in their verification. Longer, in my experience, means harder to overlook and therefore more prone to be erroneous. And I like to avoid repeating similar patters, for the same reason.
The main part is, of course, the test for a number - 0 to 255, but also making sure it doesn't allow initial zeroes (except for when it's a single one):
[1-9]?\d|1\d\d|2(5[0-5]|[0-4]\d)
Three alternations - one for sub 100: [1-9]?\d, one for 100-199: 1\d\d and finally 200-255: 2(5[0-5]|[0-4]\d).
This is preceded by a test for start of line or a dot ., and this whole expression is tested for 4 times by the appended {4}.
This complete test for four byte representations is started by testing for start of line followed by a negative look ahead to avoid addresses starting with a .: ^(?!\.), and ended with a test for end of line ($).
See some samples here at regex101.

This is what I did and it's fast and works perfectly:
function isIPv4Address(inputString) {
let regex = new RegExp(/^(([0-9]{1,3}\.){3}[0-9]{1,3})$/);
if(regex.test(inputString)){
let arInput = inputString.split(".")
for(let i of arInput){
if(i.length > 1 && i.charAt(0) === '0')
return false;
else{
if(parseInt(i) < 0 || parseInt(i) >=256)
return false;
}
}
}
else
return false;
return true;
}
Explanation: First, with the regex check that the IP format is correct. Although, the regex won't check any value ranges.
I mean, if you can use Javascript to manage regex, why not use it?. So, instead of using a crazy regex, use Regex only for checking that the format is fine and then check that each value in the octet is in the correct value range (0 to 255). Hope this helps anybody else. Peace.

And instead of
{1-3}
you should put
{1,3}

\b(?:[0-9]{1,3}\.){3}[0-9]{1,3}\b
matches 0.0.0.0 through 999.999.999.999
use if you know the seachdata does not contain invalid IP addresses
\b(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]?)\b
use to match IP numbers with accurracy - each of the 4 numbers is stored into it's own capturing group, so you can access them later

it is maybe better:
function checkIP(ip) {
var x = ip.split("."), x1, x2, x3, x4;
if (x.length == 4) {
x1 = parseInt(x[0], 10);
x2 = parseInt(x[1], 10);
x3 = parseInt(x[2], 10);
x4 = parseInt(x[3], 10);
if (isNaN(x1) || isNaN(x2) || isNaN(x3) || isNaN(x4)) {
return false;
}
if ((x1 >= 0 && x1 <= 255) && (x2 >= 0 && x2 <= 255) && (x3 >= 0 && x3 <= 255) && (x4 >= 0 && x4 <= 255)) {
return true;
}
}
return false;
}

The answers over allow leading zeros in Ip address, and that it is not correct.
For example ("123.045.067.089"should return false).
The correct way to do it like that.
function isValidIP(ipaddress) {
if (/^(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)\.(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)\.(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)\.(25[0-5]|2[0-4][0-9]|[1]?[1-9][1-9]?)$/.test(ipaddress)) {
return (true)
}
return (false) }
This function will not allow zero to lead IP addresses.

Always looking for variations, seemed to be a repetitive task so how about using forEach!
function checkIP(ip) {
//assume IP is valid to start, once false is found, always false
var test = true;
//uses forEach method to test each block of IPv4 address
ip.split('.').forEach(validateIP4);
if (!test)
alert("Invalid IP4 format\n"+ip)
else
alert("IP4 format correct\n"+ip);
function validateIP4(num, index, arr) {
//returns NaN if not an Int
item = parseInt(num, 10);
//test validates Int, 0-255 range and 4 bytes of address
// && test; at end required because this function called for each block
test = !isNaN(item) && !isNaN(num) && item >=0 && item < 256 && arr.length==4 && test;
}
}

In addition to a solution without regex:
const checkValidIpv4 = (entry) => {
const mainPipeline = [
block => !isNaN(parseInt(block, 10)),
block => parseInt(block,10) >= 0,
block => parseInt(block,10) <= 255,
block => String(block).length === 1
|| String(block).length > 1
&& String(block)[0] !== '0',
];
const blocks = entry.split(".");
if(blocks.length === 4
&& !blocks.every(block => parseInt(block, 10) === 0)) {
return blocks.every(block =>
mainPipeline.every(ckeck => ckeck(block) )
);
}
return false;
}
console.log(checkValidIpv4('0.0.0.0')); //false
console.log(checkValidIpv4('0.0.0.1')); //true
console.log(checkValidIpv4('0.01.001.0')); //false
console.log(checkValidIpv4('8.0.8.0')); //true

This should work:
function isValidIP(str) {
const arr = str.split(".").filter((el) => {
return !/^0.|\D/g.test(el);
});
return arr.filter((el) => el.length && el >= 0 && el <= 255).length === 4;
}

well I try this, I considered cases and how the entries had to be:
function isValidIP(str) {
let cong= /^(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])$/
return cong.test(str);}

A less stringent when testing the type not the validity. For example when sorting columns use this check to see which sort to use.
export const isIpAddress = (ipAddress) =>
/^((\d){1,3}\.){3}(\d){1,3}$/.test(ipAddress)
When checking for validity use this test. An even more stringent test checking that the IP 8-bit numbers are in the range 0-255:
export const isValidIpAddress = (ipAddress) =>
/^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/.test(ipAddress)

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