How to protect code from destructuring a null value in Javascript?

Question

I'm a big fan of destructuring and optional chaining proposal in Javascript.

I couldn't help but feel that when destructuring a null or undefined value it'd be really useful if the result would be undefined as it is in optional chaining instead of throwing an error.

For example:

const person = null
console.log(person?.age) // undefined

However:

const person = null
const { age } = person // runtime error

It doesn't work like this of course but is there some Babel proposal to add this feature? Is there some workaround?

Is `const { age } person` supposed to be `const { age } = person` or are you trying to convey some sort of concept with this syntax? — Nick Parsons, Sep 24 '19 at 09:57

score 6 · Accepted Answer · answered Sep 24 '19 at 09:57

6

It sounds like you might just want to use || {} to destructure an empty object in case person is is null (and happens to also apply if it's otherwise falsey too):

const person = null
const { age } = person || {};
console.log(age);

answered Sep 24 '19 at 09:57

CertainPerformance

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score 2 · Answer 2 · answered Sep 24 '19 at 09:58

2

You can use a default value using ||,

const person = null
const { age } = person || {}

console.log(age)

Short circuiting in JavaScript

answered Sep 24 '19 at 09:58

Code Maniac

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How to protect code from destructuring a null value in Javascript?

2 Answers2