Best way to alphanumeric check in JavaScript

Question

You can use this regex /^[a-z0-9]+$/i

The asker's original inclination to use str.charCodeAt(i) appears to be faster than the regular expression alternative. In my test on jsPerf the RegExp option performs 66% slower in Chrome 36 (and slightly slower in Firefox 31).

Here's a cleaned-up version of the original validation code that receives a string and returns true or false:

function isAlphaNumeric(str) {
  var code, i, len;

  for (i = 0, len = str.length; i < len; i++) {
    code = str.charCodeAt(i);
    if (!(code > 47 && code < 58) && // numeric (0-9)
        !(code > 64 && code < 91) && // upper alpha (A-Z)
        !(code > 96 && code < 123)) { // lower alpha (a-z)
      return false;
    }
  }
  return true;
};

Of course, there may be other considerations, such as readability. A one-line regular expression is definitely prettier to look at. But if you're strictly concerned with speed, you may want to consider this alternative.

Check it with a regex.

Javascript regexen don't have POSIX character classes, so you have to write character ranges manually:

if (!input_string.match(/^[0-9a-z]+$/))
  show_error_or_something()

Here ^ means beginning of string and $ means end of string, and [0-9a-z]+ means one or more of character from 0 to 9 OR from a to z.

More information on Javascript regexen here: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions

You don't need to do it one at a time. Just do a test for any that are not alpha-numeric. If one is found, the validation fails.

function validateCode(){
    var TCode = document.getElementById('TCode').value;
    if( /[^a-zA-Z0-9]/.test( TCode ) ) {
       alert('Input is not alphanumeric');
       return false;
    }
    return true;     
 }

If there's at least one match of a non alpha numeric, it will return false.