using optional chaining in array of object and destructing

Question

I have an object like this

const obj = [{a: 'a'}];

I can get it like: const { a } = obj[0]; //a const { a } = obj[0]; //a

but what if obj['x'] doesn't exist?

I tried this with optional chainning but doesn't seem to work. const { a } = obj?.[1];

Answer 1

You can use the Logical OR operator .

const { a } = obj[x] || {}

If obj does not have a property x , a will be set to undefined .

Answer 2

You are close to it. You should make sure to fallback with an empty object, for the destructing later to make sense

 const obj = [{a: 'a'}]; const { a } = obj?.[1] || {}; console.log(a)

Update

This would be even shorter if you don't use destructing in this case

 const obj = [{a: 'a'}]; const a = obj?.[1]?.a; // or: obj?.[1]?.['a'] console.log(a)

Answer 3

You could festructure with an index as computed property and a default object for a missing item of the array.

 const obj = [{ a: 'a' }], { [2]: { a } = {} } = obj; console.log(a);

Answer 4

Take a look at the following three cases, where I've used optional chaining .

You need to first check if the first element of the array ( index 0 ) exists and then check if there's a field a in it.

 // Case 1: const obj1 = [{ a: "a" }]; console.log(obj1?.[0]?.a); // "a" // Case 2: const obj2 = [{ b: "b" }]; console.log(obj2?.[0]?.a); // undefined // Case 3: const obj3 = []; console.log(obj3?.[0]?.a); // undefined