Require specific string as optional key in TypeScript interface

Question

I have a situation where I could have a number of optional, t-shirt size props added to an object. Is there a way to define a type and set it as the optional key\'s type in an i

Answer 1

A mapped type is not an interface. You can make it a type alias instead of an interface:

type TSizes = { [K in Size]?: string };

Note that I put the ? in there which makes the properties optional, as you presumably want.

Once you have such a named type you can make an interface that extends it (since all of its property names are statically known):

interface ISizes extends TSizes { }

Or you can use the built-in Partial and Record utility types to do both at the same time:

interface Sizes extends Partial> { }

Then, Sizes is an interface with optional property keys from Size whose values are of type string:

const s: Sizes = {
    s: "foo",
    m: "bar",
    xxl: "bazzz"
}

Okay, hope that helps; good luck!

Playground link to code