How the get the last element in an array items using JavaScript

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I have a list of array items like this:

const items = [
  { a: 1 },
  { b: 2 },
  { c: 3 },
]

How can I return / log the last element: { c: 3 }

Here's what I've tried so far:

let newarray = items.map((item) => {
    console.log(item);
})

console.log(newarray);
7 Answers

just log the length minus 1, nothing to do with es6:

console.log(items[items.length - 1])

I want to let you try something different:

console.log(items.slice(-1));

try this

console.log(items[items.length - 1]);

It's not required to use ES6 to perform the operation you're asking about. You could use either of the following:

/**
 * The last value in the array, `3`, is at the '2' index in the array.
 * To retrieve this value, get the length of the array, '3', and 
 * subtract 1. 
 */
const items = [1, 2, 3];
const lastItemInArray = items[items.length - 1] // => 3

or:

/**
 * Make a copy of the array by calling `slice` (to ensure we don't mutate
 * the original array) and call `pop` on the new array to return the last  
 * value from the new array.
 */
const items = [1, 2, 3];
const lastItemInArray = items.slice().pop(); // => 3

However, if you are dead set on using ES6 to retrieve this value we can leverage the spread operator (which is an ES6 feature) to retrieve the value:

/**
 * Create new array with all values in `items` array. Call `pop` on this 
 * new array to return the last value from the new array.
 *
 * NOTE: if you're using ES6 it might be a good idea to run the code
 * through Babel or some other JavaScript transpiler if you need to
 * support older browsers (IE does not support the spread operator).
 */
const items = [1, 2, 3];
const lastItemInArray = [...items].pop(); // => 3

Update - October 2021 (Chrome 97+)

Proposal for Array.prototype.findLast and Array.prototype.findLastIndex is now on Stage 3!

You can use it like this:

const items = [
  { a: 1 },
  { b: 2 },
  { c: 3 },
];

const last_element = items.findLast((item) => true);
console.log(last_element);

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