I need to find a reg ex that only allows alphanumeric. So far, everyone I try only works if the string is alphanumeric, meaning contains both a letter and a number. I just want one what would allow either and not require both.
22 Answers
/^[a-z0-9]+$/i
^ Start of string
[a-z0-9] a or b or c or ... z or 0 or 1 or ... 9
+ one or more times (change to * to allow empty string)
$ end of string
/i case-insensitive
Update (supporting universal characters)
if you need to this regexp supports universal character you can find list of unicode characters here.
for example: /^([a-zA-Z0-9\u0600-\u06FF\u0660-\u0669\u06F0-\u06F9 _.-]+)$/
this will support persian.
-
59[a-z] does not match international characters. – Eric Normand Feb 24 '12 at 22:53
-
Problem - without /g, you only end up replacing the matches before a line wrap. – Volomike Aug 13 '12 at 15:53
-
6another way would be `[^\W_]` but `[a-z0-9]`/i is a obvious way. – Vitim.us Jan 20 '13 at 05:53
-
42@Greg, I enjoy how you explain your regex-related answers. Regexes without explanations, in my opinion, are kind of useless. Because you get that one-time "yeah it works" and suddenly, when you need to change it you come right back with a different use-case, instead of actually learning what the regex *does*. Plus, you're easier to understand than regex documentation :P – Chris Cirefice Dec 19 '13 at 18:02
-
@matthewpatrickcashatt, Same here. I know right? Why is regex so cryptic? – zero_cool Dec 02 '15 at 19:56
-
This didn't work for me. Instead, I used `/[^\w]|_/g`. – christo8989 Aug 08 '16 at 23:15
-
how to check length here – R.Anandan Mar 29 '17 at 11:27
-
how to use in a function such that only these chars remain and all else is removed? – Nikhil VJ Mar 29 '18 at 14:34
-
@ksloan - Did you notice the last element. /i case-insensitive. Hope that answers your question. – Vimal Maheedharan Mar 22 '19 at 06:36
-
Sorry this answer is super dangerous, this only captures lower case letters, you'd need to do `[a-zA-Z0-9]` or just use `\W` which is all it takes – Lukas Oct 07 '19 at 11:17
-
Nice touch with allowing empty string. Super useful when validating form fields. – Oksana Romaniv Jan 29 '20 at 13:15
-
Instead of using `\u0600-\u06FF\u0660-\u0669\u06F0-\u06F9` you should use the unicode property matchers for future-proofing: `[\p{Letter}\p{Nd} _.-]+` – nightpool Dec 19 '22 at 23:19
If you wanted to return a replaced result, then this would work:
var a = 'Test123*** TEST';
var b = a.replace(/[^a-z0-9]/gi, '');
console.log(b);
This would return:
Test123TEST
Note that the gi is necessary because it means global (not just on the first match), and case-insensitive, which is why I have a-z instead of a-zA-Z. And the ^ inside the brackets means "anything not in these brackets".
WARNING: Alphanumeric is great if that's exactly what you want. But if you're using this in an international market on like a person's name or geographical area, then you need to account for unicode characters, which this won't do. For instance, if you have a name like "Âlvarö", it would make it "lvar".
UPDATE: To support unicode alphanumeric, then the REGEXP could be changed to: /[^\p{L}\p{N}]/giu
.

- 23,743
- 21
- 113
- 209
-
6This is a great answer with the explanation. Remember to allow for spaces if you need them. .replace(/[^a-z0-9\ ]/gi,'') – Joe Johnston Dec 09 '16 at 20:24
Use the word character class. The following is equivalent to a ^[a-zA-Z0-9_]+$
:
^\w+$
Explanation:
- ^ start of string
- \w any word character (A-Z, a-z, 0-9, _).
- $ end of string
Use /[^\w]|_/g
if you don't want to match the underscore.

- 2,816
- 5
- 26
- 42

- 15,703
- 8
- 51
- 58
-
24\w is actually equivalent to [a-zA-Z_0-9] so your RegEx also matches underscores [_]. – Martin Brown Dec 23 '08 at 15:36
-
almost alphanumeric is not alphanumeric at all, but thanks your ans helped me a lot – ajax333221 Sep 04 '11 at 04:14
-
True, not a strict answer to the Q as posted, but exactly what I was looking for.. – prototype Mar 28 '12 at 01:41
-
/^([a-zA-Z0-9 _-]+)$/
the above regex allows spaces in side a string and restrict special characters.It Only allows a-z, A-Z, 0-9, Space, Underscore and dash.

- 2,522
- 1
- 24
- 54

- 479
- 4
- 2
For multi-language support:
var filtered = 'Hello Привет 你好 123_456'.match(/[\p{L}\p{N}\s]/gu).join('')
console.log(filtered) // --> "Hello Привет 你好 123456"
This matches any letter, number, or space in most languages.
- [...] -> Match with conditions
- [ab] -> Match 'a' OR 'b'
- \p{L} -> Match any letter in any language
- \p{N} -> Match any number in any language
- \s -> Match a space
- /g -> Don't stop after first match
- /u -> Support unicode pattern matching

- 1,251
- 11
- 17
-
1`[a|b]` matches `a` or `|` or `b`. Do not mix with `(a|b)`! `'ab|cd'.match(/(ab|cd)/gu)` gives `['ab', 'cd']` while `'ab|cd'.match(/[ab|cd]/gu)` gives `['a', 'b', '|', 'c', 'd']`. Hence the right regex here must be `/[\p{L}\p{N}\s]/gu` (without the `|`) – Tino Nov 13 '21 at 08:59
-
^\s*([0-9a-zA-Z]*)\s*$
or, if you want a minimum of one character:
^\s*([0-9a-zA-Z]+)\s*$
Square brackets indicate a set of characters. ^ is start of input. $ is end of input (or newline, depending on your options). \s is whitespace.
The whitespace before and after is optional.
The parentheses are the grouping operator to allow you to extract the information you want.
EDIT: removed my erroneous use of the \w character set.

- 616,129
- 168
- 910
- 942
This will work
^(?=.*[a-zA-Z])(?=.*[0-9])[a-zA-Z0-9]+$
It accept only alphanumeriuc characters alone:
test cases pased :
dGgs1s23 - valid
12fUgdf - valid,
121232 - invalid,
abchfe - invalid,
abd()* - invalid,
42232^5$ - invalid
or
You can also try this one. this expression satisfied at least one number and one character and no other special characters
^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$
in angular can test like:
$scope.str = '12fUgdf';
var pattern = new RegExp('^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$');
$scope.testResult = pattern.test($scope.str);
Instead of checking for a valid alphanumeric string, you can achieve this indirectly by checking the string for any invalid characters. Do so by checking for anything that matches the complement of the valid alphanumeric string.
/[^a-z\d]/i
Here is an example:
var alphanumeric = "someStringHere";
var myRegEx = /[^a-z\d]/i;
var isValid = !(myRegEx.test(alphanumeric));
Notice the logical not operator at isValid
, since I'm testing whether the string is false, not whether it's valid.

- 369
- 3
- 11
I have string similar to Samsung Galaxy A10s 6.2-Inch (2GB,32GB ROM) Android 9.0, (13MP+2MP)+ 8MP Dual SIM 4000mAh 4G LTE Smartphone - Black (BF19)
Below is what i did:
string.replace(/[^a-zA-Z0-9 ,._-]/g, '').split(',').join('-').split(' ').join('-').toLowerCase()
Notice i allowed ,._-
then use split()
and join()
to replace ,
to -
and space to -
respectively.
I ended up getting something like this:
samsung-galaxy-a10s-6.2-inch-2gb-32gb-rom-android-9.0-13mp-2mp-8mp-dual-sim-4000mah-4g-lte-smartphone-black-bf19-20
which is what i wanted.
There might be a better solution but this is what i found working fine for me.

- 1,119
- 15
- 20
Extend the string prototype to use throughout your project
String.prototype.alphaNumeric = function() {
return this.replace(/[^a-z0-9]/gi,'');
}
Usage:
"I don't know what to say?".alphaNumeric();
//Idontknowwhattosay

- 677
- 6
- 11
Even better than Gayan Dissanayake
pointed out.
/^[-\w\s]+$/
Now ^[a-zA-Z0-9]+$
can be represented as ^\w+$
You may want to use \s instead of space. Note that \s takes care of whitespace and not only one space character.

- 4,958
- 8
- 40
- 56
Input these code to your SCRATCHPAD and see the action.
var str=String("Blah-Blah1_2,oo0.01&zz%kick").replace(/[^\w-]/ig, '');

- 552
- 5
- 8
JAVASCRIPT to accept only NUMBERS, ALPHABETS and SPECIAL CHARECTERS
document.getElementById("onlynumbers").onkeypress = function (e) {
onlyNumbers(e.key, e)
};
document.getElementById("onlyalpha").onkeypress = function (e) {
onlyAlpha(e.key, e)
};
document.getElementById("speclchar").onkeypress = function (e) {
speclChar(e.key, e)
};
function onlyNumbers(key, e) {
var letters = /^[0-9]/g; //g means global
if (!(key).match(letters)) e.preventDefault();
}
function onlyAlpha(key, e) {
var letters = /^[a-z]/gi; //i means ignorecase
if (!(key).match(letters)) e.preventDefault();
}
function speclChar(key, e) {
var letters = /^[0-9a-z]/gi;
if ((key).match(letters)) e.preventDefault();
}
<html>
<head></head>
<body>
Enter Only Numbers:
<input id="onlynumbers" type="text">
<br><br>
Enter Only Alphabets:
<input id="onlyalpha" type="text" >
<br><br>
Enter other than Alphabets and numbers like special characters:
<input id="speclchar" type="text" >
</body>
</html>

- 59
- 4
A little bit late, but this worked for me:
/[^a-z A-Z 0-9]+/g
a-z : anything from a to z.
A-Z : anything from A to Z (upper case).
0-9 : any number from 0 to 9.
It will allow anything inside square brackets, so let's say you want to allow any other character, for example, "/" and "#", the regex would be something like this:
/[^a-z A-Z 0-9 / #]+/g
This site will help you to test your regex before coding. https://regex101.com/
Feel free to modify and add anything you want into the brackets. Regards :)

- 66
- 4
It seems like many users have noticed this these regular expressions will almost certainly fail unless we are strictly working in English. But I think there is an easy way forward that would not be so limited.
- make a copy of your string in all UPPERCASE
- make a second copy in all lowercase
Any characters that match in those strings are definitely not alphabetic in nature.
let copy1 = originalString.toUpperCase();
let copy2 = originalString.toLowerCase();
for(let i=0; i<originalString.length; i++) {
let bIsAlphabetic = (copy1[i] != copy2[i]);
}
Optionally, you can also detect numerics by just looking for digits 0 to 9.

- 157
- 1
- 11
Try this... Replace you field ID with #name... a-z(a to z), A-Z(A to Z), 0-9(0 to 9)
jQuery(document).ready(function($){
$('#name').keypress(function (e) {
var regex = new RegExp("^[a-zA-Z0-9\s]+$");
var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
if (regex.test(str)) {
return true;
}
e.preventDefault();
return false;
});
});

- 1,770
- 21
- 28
Save this constant
const letters = /^[a-zA-Z0-9]+$/
now, for checking part use .match()
const string = 'Hey there...' // get string from a keyup listner
let id = ''
// iterate through each letters
for (var i = 0; i < string.length; i++) {
if (string[i].match(letters) ) {
id += string[i]
} else {
// In case you want to replace with something else
id += '-'
}
}
return id

- 745
- 8
- 13
Alphanumeric with case sensitive:
if (/^[a-zA-Z0-9]+$/.test("SoS007")) {
alert("match")
}

- 2,369
- 1
- 27
- 37
Also if you were looking for just Alphabetical characters, you can use the following regular expression:
/[^a-zA-Z]/gi
Sample code in typescript:
let samplestring = "!#!&34!# Alphabet !!535!!! is safe"
let regex = new RegExp(/[^a-zA-Z]/gi);
let res = samplestring.replace(regex,'');
console.log(res);
Note: if you are curious about RegEx syntax, visit regexr and either use the cheat-sheet or play with regular expressions.
Edit: alphanumeric --> alphabetical

- 19
- 4
-
2
-
Correct, to handle numbers you could just add 0-9 or whatever you wish to cover after A-Z, my apologies I meant to write alphabetical rather than alphanumeric – Tom Zdanowski Apr 29 '21 at 18:55
-
@TomZdanowski it would be helpful if you could edit your answer so that it completely responds to the question, otherwise it's incorrect (and some uniform formatting wouldn't hurt either) – derekbaker783 Jan 23 '22 at 02:38
Only accept numbers and letters (No Space)
function onlyAlphanumeric(str){
str.value=str.value.replace(/\s/g, "");//No Space
str.value=str.value.replace(/[^a-zA-Z0-9 ]/g, "");
}
<div>Only accept numbers and letters </div>
<input type="text" onKeyUp="onlyAlphanumeric(this);" >

- 3,067
- 1
- 34
- 33
Here is the way to check:
/**
* If the string contains only letters and numbers both then return true, otherwise false.
* @param string
* @returns boolean
*/
export const isOnlyAlphaNumeric = (string: string) => {
return /^(?=.*[a-zA-Z])(?=.*[0-9])[a-zA-Z0-9]+$/.test(string);
}

- 8,783
- 6
- 58
- 79
Jquery to accept only NUMBERS, ALPHABETS and SPECIAL CHARECTERS
<html>
<head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
Enter Only Numbers:
<input type="text" id="onlynumbers">
<br><br>
Enter Only Alphabets:
<input type="text" id="onlyalpha">
<br><br>
Enter other than Alphabets and numbers like special characters:
<input type="text" id="speclchar">
<script>
$('#onlynumbers').keypress(function(e) {
var letters=/^[0-9]/g; //g means global
if(!(e.key).match(letters)) e.preventDefault();
});
$('#onlyalpha').keypress(function(e) {
var letters=/^[a-z]/gi; //i means ignorecase
if(!(e.key).match(letters)) e.preventDefault();
});
$('#speclchar').keypress(function(e) {
var letters=/^[0-9a-z]/gi;
if((e.key).match(letters)) e.preventDefault();
});
</script>
</body>
</html>
**JQUERY to accept only NUMBERS , ALPHABETS and SPECIAL CHARACTERS **
<!DOCTYPE html>
$('#onlynumbers').keypress(function(e) {
var letters=/^[0-9]/g; //g means global
if(!(e.key).match(letters)) e.preventDefault();
});
$('#onlyalpha').keypress(function(e) {
var letters=/^[a-z]/gi; //i means ignorecase
if(!(e.key).match(letters)) e.preventDefault();
});
$('#speclchar').keypress(function(e) {
var letters=/^[0-9a-z]/gi;
if((e.key).match(letters)) e.preventDefault();
});
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js">
Enter Only Numbers:
Enter Only Alphabets:
Enter other than Alphabets and numbers like special characters:
</body>
</html>

- 59
- 4
-
3
-
Yeah. Thanks above code is working you can check by clicking run code snippet. – Pranav Bhat Jul 28 '19 at 12:14
-