I want a regex that would match this range: 0.5 - 24, but not this: 0,5 or 22,5. For now I have this one:
/^(([0-9]|1[0-9]|2[0-3])([^,])(\.(0|5)))$/
but for some reason it matches this 22,5 even though I negated the ,. Thanks!
You just need to only allow the dot and make a last capture group optional.
Like this:
/^([0-9]|1[0-9]|2[0-3])(\.[05])?$/
(\.[05])? <- this allows, by the ? symbol, one or zero ocurrences of the second capture group.
In terms of regular expressions, this will suit your needs.
^((0\.[5-9])|(([1-9](\.[5-9])?)|(1[0-9](\.[5-9])?)|(2[0-3](\.[5-9])?)|24))$
You can test it here
However, if what you want is to check a range, I would advise you to use regular expressions just to check if the number is valid, and then, convert it and use plain javascript for your validation.
var validator = function(value) {
var regexp = /^-?\d+(\.\d{0,2})?$/;
var isNumber = regexp.test(value);
if (isNumber) {
var parsed = parseFloat(value);
if (0.5 <= parsed && parsed <= 24)
console.log("The number " + parsed + " is within range");
else
console.log("The number " + parsed + " is NOT within range");
} else {
console.log("The value " + value + " is NOT even a valid decimal value");
}
}
validator("-3");
validator("19.6");
validator("22,5");
In javascript if I have something like
string.replace(new RegExp(regex, "ig"), " ")
this replaces all found regexes with a single space. But how would I do it if I wanted to replace all found regexes with spaces that matched in length?
so if regex was \d+, and the string was
"123hello4567"
it changes to
" hello "
Thanks
The replacement argument (2nd) to .replace can be a function - this function is called in turn with every matching part as the first argument
knowing the length of the matching part, you can return the same number of spaces as the replacement value
In the code below I use . as a replacement value to easily illustrate the code
Note: this uses String#repeat, which is not available in IE11 (but then, neither are arrow functions) but you can always use a polyfill and a transpiler
let regex = "\\d+";
console.log("123hello4567".replace(new RegExp(regex, "ig"), m => '.'.repeat(m.length)));
Internet Exploder friendly version
var regex = "\\d+";
console.log("123hello4567".replace(new RegExp(regex, "ig"), function (m) {
return Array(m.length+1).join('.');
}));
thanks to #nnnnnn for the shorter IE friendly version
"123hello4567".replace(new RegExp(/[\d]/, "ig"), " ")
1 => " "
2 => " "
3 => " "
" hello "
"123hello4567".replace(new RegExp(/[\d]+/, "ig"), " ")
123 => " "
4567 => " "
" hello "
If you just want to replace every digit with a space, keep it simple:
var str = "123hello4567";
var res = str.replace(/\d/g,' ');
" hello "
This answers your example, but not exactly your question. What if the regex could match on different numbers of spaces depending on the string, or it isn't as simple as /d more than once? You could do something like this:
var str = "123hello456789goodbye12and456hello12345678again123";
var regex = /(\d+)/;
var match = regex.exec(str);
while (match != null) {
// Create string of spaces of same length
var replaceSpaces = match[0].replace(/./g,' ');
str = str.replace(regex, replaceSpaces);
match = regex.exec(str);
}
" hello goodbye and hello again "
Which will loop through executing the regex (instead of using /g for global).
Performance wise this could likely be sped up by creating a new string of spaces with the length the same length as match[0]. This would remove the regex replace within the loop. If performance isn't a high priority, this should work fine.
Let say, I get the following using var content = this.innerHTML:
w here </div>
Using indexOf (or other ways), I want to check for the first position that has either "Space", "<" or " ".
In this case, it will be 1 (after "w").
What I am confused about is how do I check for the very first position that has either one of these three choices? Do I use Do...while to check for individual "options"?
You're probably looking for a Regular Expression (Regex) and the String#search method. Regex is a bit much to learn all at once, but I'll explain this example code.
You can use square brackets to denote a set of characters, so for example [ <] says "match either a space or a less-than sign."
You can use the pipe | to separate possibilities if you want to match one pattern or another, and that's how to account for matching a non-breaking space HTML entity.
var string = 'w here </div>',
index = string.search(/[ <]| /)
console.log(index) //=> 1
You can use a regular expression with alternations (|), which means "match one of these things". That will also tell you what you found, if that's useful:
function check(str) {
var m = / |<| /.exec(str);
if (!m) {
console.log("Not found in '" + str + "'");
return;
}
console.log("'" + m[0] + "' found at index " + m.index + " in '" + str + "'");
}
check("w here </div>");
check("where </div>");
check("where</div>");
I'm looking into using a jQuery password strength indicator and have found one that looks suitable.
It increases the password strength score if special characters are detected:
if (password.match(/(.*[!,#,#,$,%,^,&,*,?,_,~].*[!,#,#,$,%,^,&,*,?,_,~])/)){ score += 5 ;}
However I'd like to be able to specify additional special characters and because these lists of special characters are used in several places, I'd like to only specify the list once:
list = array(!,#,#,$,%,^,&,*,?,_,~,[,],{,},(,));
if (password.match(/(.*[list].*[list])/)){ score += 5 ;}
Is this possible?
You can use strings:
var special = "!##$%^&*?_~[]{}()".split('').join('\\');
if (password.match(new RegExp("(.*[" + special + "].*[" + special + "])")))...
(The join-with-backslashes escapes the special characters so they are treated literally by the regex engine.)
Yes, if you use the RegExp() constructor, you can pass in a string as regexp.
var list = ['\\!', '\\#', '\\#', '\\%'];
var reg = new RegExp('(.*['+ list.join(',') + '].*['+ list.join(',') +'])');
if (reg.test("MySuperPassword!#_123")) {
score += 5;
}
You do not need to separate chars by , in regex:
var list = "[\\!##\\$%\\^&\\*\\?_~]";
var your_regex = new RegExp(".*" + list + ".*" + list);
if (your_regex.test(password)){
score += 5;
}
Why would you need a regex?
var list = ['!','#','#','$','%','^','&','*','?','_','~','[',']','{','}','(',')'],
score = 0;
for (var i=list.length;i--;) {
if ( password.indexOf(list[i]) ) score++;
}
FIDDLE
I'm looking to reformat (replace, not validate - there are many references for validating) a phone number for display in Javascript. Here's an example of some of the data:
123 4567890
(123) 456-7890
(123)456-7890
123 456 7890
123.456.7890
(blank/null)
1234567890
Is there an easy way to use a regular expression to do this? I'm looking for the best way to do this. Is there a better way?
I want to reformat the number to the following: (123) 456-7890
Assuming you want the format "(123) 456-7890":
function formatPhoneNumber(phoneNumberString) {
var cleaned = ('' + phoneNumberString).replace(/\D/g, '');
var match = cleaned.match(/^(\d{3})(\d{3})(\d{4})$/);
if (match) {
return '(' + match[1] + ') ' + match[2] + '-' + match[3];
}
return null;
}
Here's a version that allows the optional +1 international code:
function formatPhoneNumber(phoneNumberString) {
var cleaned = ('' + phoneNumberString).replace(/\D/g, '');
var match = cleaned.match(/^(1|)?(\d{3})(\d{3})(\d{4})$/);
if (match) {
var intlCode = (match[1] ? '+1 ' : '');
return [intlCode, '(', match[2], ') ', match[3], '-', match[4]].join('');
}
return null;
}
formatPhoneNumber('+12345678900') // => "+1 (234) 567-8900"
formatPhoneNumber('2345678900') // => "(234) 567-8900"
Possible solution:
function normalize(phone) {
//normalize string and remove all unnecessary characters
phone = phone.replace(/[^\d]/g, "");
//check if number length equals to 10
if (phone.length == 10) {
//reformat and return phone number
return phone.replace(/(\d{3})(\d{3})(\d{4})/, "($1) $2-$3");
}
return null;
}
var phone = '(123)4567890';
phone = normalize(phone); //(123) 456-7890
var x = '301.474.4062';
x = x.replace(/\D+/g, '')
.replace(/(\d{3})(\d{3})(\d{4})/, '($1) $2-$3');
alert(x);
This answer borrows from maerics' answer. It differs primarily in that it accepts partially entered phone numbers and formats the parts that have been entered.
phone = value.replace(/\D/g, '');
const match = phone.match(/^(\d{1,3})(\d{0,3})(\d{0,4})$/);
if (match) {
phone = `${match[1]}${match[2] ? ' ' : ''}${match[2]}${match[3] ? '-' : ''}${match[3]}`;
}
return phone
I'm using this function to format US numbers.
function formatUsPhone(phone) {
var phoneTest = new RegExp(/^((\+1)|1)? ?\(?(\d{3})\)?[ .-]?(\d{3})[ .-]?(\d{4})( ?(ext\.? ?|x)(\d*))?$/);
phone = phone.trim();
var results = phoneTest.exec(phone);
if (results !== null && results.length > 8) {
return "(" + results[3] + ") " + results[4] + "-" + results[5] + (typeof results[8] !== "undefined" ? " x" + results[8] : "");
}
else {
return phone;
}
}
It accepts almost all imaginable ways of writing a US phone number. The result is formatted to a standard form of (987) 654-3210 x123
thinking backwards
Take the last digits only (up to 10) ignoring first "1".
function formatUSNumber(entry = '') {
const match = entry
.replace(/\D+/g, '').replace(/^1/, '')
.match(/([^\d]*\d[^\d]*){1,10}$/)[0]
const part1 = match.length > 2 ? `(${match.substring(0,3)})` : match
const part2 = match.length > 3 ? ` ${match.substring(3, 6)}` : ''
const part3 = match.length > 6 ? `-${match.substring(6, 10)}` : ''
return `${part1}${part2}${part3}`
}
example input / output as you type
formatUSNumber('+1333')
// (333)
formatUSNumber('333')
// (333)
formatUSNumber('333444')
// (333) 444
formatUSNumber('3334445555')
// (333) 444-5555
2021
libphonenumber-js
Example
import parsePhoneNumber from 'libphonenumber-js'
const phoneNumber = parsePhoneNumber('+12133734253')
phoneNumber.formatInternational() === '+1 213 373 4253'
phoneNumber.formatNational() === '(213) 373-4253'
phoneNumber.getURI() === 'tel:+12133734253'
Based on David Baucum's answer - here is a version that trys to improve auto-replacement "as you type" for example in a React onChange event handler:
function formatPhoneNumber(phoneNumber) {
const cleanNum = phoneNumber.toString().replace(/\D/g, '');
const match = cleanNum.match(/^(\d{3})(\d{0,3})(\d{0,4})$/);
if (match) {
return '(' + match[1] + ') ' + (match[2] ? match[2] + "-" : "") + match[3];
}
return cleanNum;
}
//...
onChange={e => setPhoneNum(formatPhoneNumber(e.target.value))}
It will insert (###) as soon as there are 3 numbers and then it will keep following the RegEx until it looks like this (###) ###-####
I've extended David Baucum's answer to include support for extensions up to 4 digits in length. It also includes the parentheses requested in the original question. This formatting will work as you type in the field.
phone = phone.replace(/\D/g, '');
const match = phone.match(/^(\d{1,3})(\d{0,3})(\d{0,4})(\d{0,4})$/);
if (match) {
phone = `(${match[1]}${match[2] ? ') ' : ''}${match[2]}${match[3] ? '-' : ''}${match[3]}${match[4] ? ' x' : ''}${match[4]}`;
}
return phone;
Almost all of these have issues when the user tries to backspace over the delimiters, particularly from the middle of the string.
Here's a jquery solution that handles that, and also makes sure the cursor stays in the right place as you edit:
//format text input as phone number (nnn) nnn-nnnn
$('.myPhoneField').on('input', function (e){
var $phoneField = e.target;
var cursorPosition = $phoneField.selectionStart;
var numericString = $phoneField.value.replace(/\D/g, '').substring(0, 10);
// let user backspace over the '-'
if (cursorPosition === 9 && numericString.length > 6) return;
// let user backspace over the ') '
if (cursorPosition === 5 && numericString.length > 3) return;
if (cursorPosition === 4 && numericString.length > 3) return;
var match = numericString.match(/^(\d{1,3})(\d{0,3})(\d{0,4})$/);
if (match) {
var newVal = '(' + match[1];
newVal += match[2] ? ') ' + match[2] : '';
newVal += match[3] ? '-' + match[3] : '';
// to help us put the cursor back in the right place
var delta = newVal.length - Math.min($phoneField.value.length, 14);
$phoneField.value = newVal;
$phoneField.selectionEnd = cursorPosition + delta;
} else {
$phoneField.value = '';
}
})
var numbers = "(123) 456-7890".replace(/[^\d]/g, ""); //This strips all characters that aren't digits
if (numbers.length != 10) //wrong format
//handle error
var phone = "(" + numbers.substr(0, 3) + ") " + numbers.substr(3, 3) + "-" + numbers.substr(6); //Create format with substrings
Here is one that will accept both phone numbers and phone numbers with extensions.
function phoneNumber(tel) {
var toString = String(tel),
phoneNumber = toString.replace(/[^0-9]/g, ""),
countArrayStr = phoneNumber.split(""),
numberVar = countArrayStr.length,
closeStr = countArrayStr.join("");
if (numberVar == 10) {
var phone = closeStr.replace(/(\d{3})(\d{3})(\d{4})/, "$1.$2.$3"); // Change number symbols here for numbers 10 digits in length. Just change the periods to what ever is needed.
} else if (numberVar > 10) {
var howMany = closeStr.length,
subtract = (10 - howMany),
phoneBeginning = closeStr.slice(0, subtract),
phoneExtention = closeStr.slice(subtract),
disX = "x", // Change the extension symbol here
phoneBeginningReplace = phoneBeginning.replace(/(\d{3})(\d{3})(\d{4})/, "$1.$2.$3"), // Change number symbols here for numbers greater than 10 digits in length. Just change the periods and to what ever is needed.
array = [phoneBeginningReplace, disX, phoneExtention],
afterarray = array.splice(1, 0, " "),
phone = array.join("");
} else {
var phone = "invalid number US number";
}
return phone;
}
phoneNumber("1234567891"); // Your phone number here
For all international Phone numbers with country code upto 3 digits, we can change the original answer a little bit as below.
For first match instead of looking for '1' we should look for 1-3 digits.
export const formatPhoneNumber = (phoneNumberString) => {
var cleaned = ('' + phoneNumberString).replace(/\D/g, '');
var match = cleaned.match(/^(\d{1,3}|)?(\d{3})(\d{3})(\d{4})$/);
if (match) {
var intlCode = (match[1] ? `+${match[1]} ` : '');
return [intlCode, '(', match[2], ') ', match[3], '-', match[4]].join('');
}
return null;
}
console.log( formatPhoneNumber('16464765278') )//+1 (646) 476-5278
console.log( formatPhoneNumber('+2549114765278')) //+254 (911) 476-5278
console.log( formatPhoneNumber('929876543210') )//+92 (987) 654-3210
Fulfils my requirement.
For US Phone Numbers
/^\(?(\d{3})\)?[- ]?(\d{3})[- ]?(\d{4})$/
Let’s divide this regular expression in smaller fragments to make is easy to understand.
/^\(?: Means that the phone number may begin with an optional (.
(\d{3}): After the optional ( there must be 3 numeric digits. If the phone number does not have a (, it must start with 3 digits. E.g. (308 or 308.
\)?: Means that the phone number can have an optional ) after first 3 digits.
[- ]?: Next the phone number can have an optional hyphen (-) after ) if present or after first 3 digits.
(\d{3}): Then there must be 3 more numeric digits. E.g (308)-135 or 308-135 or 308135
[- ]?: After the second set of 3 digits the phone number can have another optional hyphen (-). E.g (308)-135- or 308-135- or 308135-
(\d{4})$/: Finally, the phone number must end with four digits. E.g (308)-135-7895 or 308-135-7895 or 308135-7895 or 3081357895.
Reference :
http://www.zparacha.com/phone_number_regex/
You can use this functions to check valid phone numbers and normalize them:
let formatPhone = (dirtyNumber) => {
return dirtyNumber.replace(/\D+/g, '').replace(/(\d{3})(\d{3})(\d{4})/, '($1) $2-$3');
}
let isPhone = (phone) => {
//normalize string and remove all unnecessary characters
phone = phone.replace(/\D+/g, '');
return phone.length == 10? true : false;
}
The solutions above are superior, especially if using Java, and encountering more numbers with more than 10 digits such as the international code prefix or additional extension numbers. This solution is basic (I'm a beginner in the regex world) and designed with US Phone numbers in mind and is only useful for strings with just 10 numbers with perhaps some formatting characters, or perhaps no formatting characters at all (just 10 numbers). As such I would recomend this solution only for semi-automatic applications. I Personally prefer to store numbers as just 10 numbers without formatting characters, but also want to be able to convert or clean phone numbers to the standard format normal people and apps/phones will recognize instantly at will.
I came across this post looking for something I could use with a text cleaner app that has PCRE Regex capabilities (but no java functions). I will post this here for people who could use a simple pure Regex solution that could work in a variety of text editors, cleaners, expanders, or even some clipboard managers. I personally use Sublime and TextSoap. This solution was made for Text Soap as it lives in the menu bar and provides a drop-down menu where you can trigger text manipulation actions on what is selected by the cursor or what's in the clipboard.
My approach is essentially two substitution/search and replace regexes. Each substitution search and replace involves two regexes, one for search and one for replace.
Substitution/ Search & Replace #1
The first substitution/ search & replace strips non-numeric numbers from an otherwise 10-digit number to a 10-digit string.
First Substitution/ Search Regex: \D
This search string matches all characters that is not a digit.
First Substitution/ Replace Regex: "" (nothing, not even a space)
Leave the substitute field completely blank, no white space should exist including spaces. This will result in all matched non-digit characters being deleted. You should have gone in with 10 digits + formatting characters prior this operation and come out with 10 digits sans formatting characters.
Substitution/ Search & Replace #2
The second substitution/search and replace search part of the operation captures groups for area code $1, a capture group for the second set of three numbers $2, and the last capture group for the last set of four numbers $3. The regex for the substitute portion of the operation inserts US phone number formatting in between the captured group of digits.
Second Substitution/ Search Regex: (\d{3})(\d{3})(\d{4})
Second Substitution/ Replace Regex: \($1\) $2\-$3
The backslash \ escapes the special characters (, ) , (<-whitespace), and - since we are inserting them between our captured numbers in capture groups $1, $2, & $3 for US phone number formatting purposes.
In TextSoap I created a custom cleaner that includes the two substitution operation actions, so in practice it feels identical to executing a script. I'm sure this solution could be improved but I expect complexity to go up quite a bit. An improved version of this solution is welcomed as a learning experience if anyone wants to add to this.