RegEx for Javascript to allow only alphanumeric

/^[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.

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".

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.