Best way to alphanumeric check in JavaScript

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