# 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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