Determine if JavaScript value is an "integer"?

Here's a polyfill for the Number predicate functions:

"use strict";

Number.isNaN = Number.isNaN ||
    n => n !== n; // only NaN

Number.isNumeric = Number.isNumeric ||
    n => n === +n; // all numbers excluding NaN

Number.isFinite = Number.isFinite ||
    n => n === +n               // all numbers excluding NaN
      && n >= Number.MIN_VALUE  // and -Infinity
      && n <= Number.MAX_VALUE; // and +Infinity

Number.isInteger = Number.isInteger ||
    n => n === +n              // all numbers excluding NaN
      && n >= Number.MIN_VALUE // and -Infinity
      && n <= Number.MAX_VALUE // and +Infinity
      && !(n % 1);             // and non-whole numbers

Number.isSafeInteger = Number.isSafeInteger ||
    n => n === +n                     // all numbers excluding NaN
      && n >= Number.MIN_SAFE_INTEGER // and small unsafe numbers
      && n <= Number.MAX_SAFE_INTEGER // and big unsafe numbers
      && !(n % 1);                    // and non-whole numbers

All major browsers support these functions, except isNumeric, which is not in the specification because I made it up. Hence, you can reduce the size of this polyfill:

"use strict";

Number.isNumeric = Number.isNumeric ||
    n => n === +n; // all numbers excluding NaN

Alternatively, just inline the expression n === +n manually.

Use jQuery's IsNumeric method.

http://api.jquery.com/jQuery.isNumeric/

if ($.isNumeric(id)) {
   //it's numeric
}

CORRECTION: that would not ensure an integer. This would:

if ( (id+"").match(/^\d+$/) ) {
   //it's all digits
}

That, of course, doesn't use jQuery, but I assume jQuery isn't actually mandatory as long as the solution works

Try this:

if(Math.floor(id) == id && $.isNumeric(id)) 
  alert('yes its an int!');

$.isNumeric(id) checks whether it's numeric or not
Math.floor(id) == id will then determine if it's really in integer value and not a float. If it's a float parsing it to int will give a different result than the original value. If it's int both will be the same.