Destructuring in Typescript with undefined values

Question

I have 2 objects:

const a = {
    foo: "foo",
    bar: "bar",
}

const b = {
    foo: "fooooo",
}

I want to use destructuring in a method with default undefined values, like so:

const c = a or b; // I don't know which one

Then I want to do:

const { foo, bar } = c;

And I want that

foo = "fooooo" and bar = undefined or
foo = "foo" and bar = "bar"

How can I accomplish that with typescript?

Madara's Ghost · Accepted Answer · 2018-06-05 20:50:27Z

3

TypeScript won't be smart enough to deduct that {foo: string, bar: string} | {foo: string} can be written as {foo: string, bar?: string}, so you'll need to type c yourself as follows:

const c: { foo: string, bar?: string } = Math.random() > .5 ? a : b; // doesn't matter which
const { foo, bar } = c;

answered Jun 5, 2018 at 20:50

Madara's Ghost

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2 Comments

Gustavo Lopes Over a year ago

The problem with this approach is that I don't want c to have the bar property. That's why I wanted the destructuring to return undefined, so c would still not have the bar property. I guess I'll have to create a wrapper function for that. Thank you

Madara's Ghost Over a year ago

What do you mean? c might have the bar property whether you want to or not, if you set c to be a, it's going to have it... If not, it will get undefined.

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