How to check if a string is a natural number?

Here is my solution:

function isNaturalNumber(n) {
    n = n.toString(); // force the value incase it is not
    var n1 = Math.abs(n),
        n2 = parseInt(n, 10);
    return !isNaN(n1) && n2 === n1 && n1.toString() === n;
}

Here is the demo:

var tests = [
        '0',
        '1',
        '-1',
        '-1.1',
        '1.1',
        '12abc123',
        '+42',
        '0xFF',
        '5e3'
    ];

function isNaturalNumber(n) {
    n = n.toString(); // force the value incase it is not
    var n1 = Math.abs(n),
        n2 = parseInt(n, 10);
    return !isNaN(n1) && n2 === n1 && n1.toString() === n;
}

console.log(tests.map(isNaturalNumber));

here is the output:

[true, true, false, false, false, false, false, false, false]

DEMO: http://jsfiddle.net/rlemon/zN6j3/1

Note: this is not a true natural number, however I understood it that the OP did not want a real natural number. Here is the solution for real natural numbers:

function nat(n) {
    return n >= 0 && Math.floor(n) === +n;
}

http://jsfiddle.net/KJcKJ/

provided by @BenjaminGruenbaum

Use a regular expression

function isNaturalNumber (str) {
    var pattern = /^(0|([1-9]\d*))$/;
    return pattern.test(str);
}

The function will return either true or false so you can do a check based on that.

if(isNaturalNumber(number)){ 
   // Do something if the number is natural
}else{
   // Do something if it's not natural
}

Source: http://www.codingforums.com/showthread.php?t=148668