Testing whether a value is odd or even

Use modulus:

function isEven(n) {
   return n % 2 == 0;
}

function isOdd(n) {
   return Math.abs(n % 2) == 1;
}

You can check that any value in Javascript can be coerced to a number with:

Number.isFinite(parseFloat(n))

This check should preferably be done outside the isEven and isOdd functions, so you don't have to duplicate error handling in both functions.

I prefer using a bit test:

if(i & 1)
{
    // ODD
}
else
{
    // EVEN
}

This tests whether the first bit is on which signifies an odd number.

Note: there are also negative numbers.

function isOddInteger(n)
{
   return isInteger(n) && (n % 2 !== 0);
}

where

function isInteger(n)
{
   return n === parseInt(n, 10);
}

How about the following? I only tested this in IE, but it was quite happy to handle strings representing numbers of any length, actual numbers that were integers or floats, and both functions returned false when passed a boolean, undefined, null, an array or an object. (Up to you whether you want to ignore leading or trailing blanks when a string is passed in - I've assumed they are not ignored and cause both functions to return false.)

function isEven(n) {
   return /^-?\d*[02468]$/.test(n);
}

function isOdd(n) {
   return /^-?\d*[13579]$/.test(n);
}

To complete Robert Brisita's bit test .

if ( ~i & 1 ) {
    // Even
}

var isEven = function(number) {
    // Your code goes here!
    if (number % 2 == 0){
       return(true);
    }
    else{
       return(false);    
    }
};

A simple modification/improvement of Steve Mayne answer!

function isEvenOrOdd(n){
    if(n === parseFloat(n)){
        return isNumber(n) && (n % 2 == 0);
    }
    return false;
}

Note: Returns false if invalid!

We just need one line of code for this!

Here a newer and alternative way to do this, using the new ES6 syntax for JS functions, and the one-line syntax for the if-else statement call:

const isEven = num => ((num % 2) == 0) ? true : false;

alert(isEven(8));  //true
alert(isEven(9));  //false
alert(isEven(-8)); //true

Different way:

var isEven = function(number) {
  // Your code goes here!
  if (((number/2) - Math.floor(number/2)) === 0) {return true;} else {return false;};
};

isEven(69)

Otherway using strings because why not

function isEven(__num){
    return String(__num/2).indexOf('.') === -1;
}

if (testNum == 0);
else if (testNum % 2  == 0);
else if ((testNum % 2) != 0 );

Maybe this? if(ourNumber % 2 !== 0)

To test whether or not you have a odd or even number, this also works.

const comapare = x => integer(checkNumber(x));

function checkNumber (x) {
   if (x % 2 == 0) {
       return true;
   } 
   else if (x % 2 != 0) {
       return false;
    }
}

function integer (x) {
   if (x) {
       console.log('even');
   } 
   else {
       console.log('odd');
    }
}

Using modern javascript style:

const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")

const isOdd  = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = n=> isOdd(+n+1)

Simple way to test whether a number is even using JavaScript

function isEven(x) {
//return true if even
if (x % 2 === 0) {
    return true;
}
//return false otherwise
else {
    return false
}
}
// or much simpler
function isEven(x) {
return x % 2 === 0;
}

This one is more simple!

  var num = 3 //instead get your value here
  var aa = ["Even", "Odd"];

  alert(aa[num % 2]);

if (i % 2) {
return odd numbers
}

if (i % 2 - 1) {
return even numbers
}

Why not just do this:

    function oddOrEven(num){
        if(num % 2 == 0)
            return "even";
        return "odd";
    }
    oddOrEven(num);

var num = someNumber
    isEven;
parseInt(num/2) === num/2 ? isEven = true : isEven = false;

A few

x % 2 == 0; // Check if even

!(x & 1); // bitmask the value with 1 then invert.

((x >> 1) << 1) == x; // divide value by 2 then multiply again and check against original value

~x&1; // flip the bits and bitmask

function isEven(n) {return parseInt(n)%2===0?true:parseInt(n)===0?true:false}

when 0/even wanted but

isEven(0) //true
isEven(1) //false
isEven(2) //true
isEven(142856) //true
isEven(142856.142857)//true
isEven(142857.1457)//false

?