How to Return All Even Numbers in a JavaScript Array?

You can find all even numbers in a JavaScript array by:

The check for even numbers can be done using the remainder operator (%), where an even number would return true if n % 2 === 0 (where n is an integer).

Using Array.prototype.filter()

To get all even numbers in an array of integers using the Array.prototype.filter() method, you can do the following:

// ES5+
const numbers = [-5, -2, -1, 0, 1, 3, 4, 7];
const evenNumbers = numbers.filter(function (number) {
    return number % 2 === 0;
});

console.log(evenNumbers); // [-2, 0, 4]

This would return a new array containing all even numbers. In ES6+, you can shorten this to a one-liner using the arrow function syntax:

// ES6+
const numbers = [-5, -2, -1, 0, 1, 3, 4, 7];
const evenNumbers = numbers.filter((number) => number % 2 === 0);

console.log(evenNumbers); // [-2, 0, 4]

When no matches are found, an empty array is returned:

// ES6+
const numbers = [-5, -1, 1, 3, 7];
const evenNumbers = numbers.filter((number) => number % 2 === 0);

console.log(evenNumbers); // []

Using a Loop

You can loop over an array of integers and create a new array to which you add all even numbers to, for example, like so:

const numbers = [-5, -2, -1, 0, 1, 3, 4, 7];
const evenNumbers = [];

for (let i = 0; i < numbers.length; i++) {
    if (numbers[i] % 2 === 0) {
        evenNumbers.push(numbers[i]);
    }
}

console.log(evenNumbers); // [-2, 0, 4]

This would return a new array containing all even numbers, and when no match is found, an empty array is returned.


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