How to get a literal type from a function argument for a computed property name?

Question

please consider the following code:

const fn = (name: string) => {
  return { [name]: "some txt" };
};

const res = fn("books"); // books or any other string

TS recognizes res as the following type:

const res: {
  [x: string]: string;
}

I'd like TS to know that res has a property books

const res: {
  books: string;
}

I tried many things but nothing seems to work. Is it possible at all? Is it a known issue?

Answer 1

You have to create a generic function like this:

const fn = <T extends string>(name: T): { [key in T]: string } => {
  return { [name]: "some txt" } as any;
};

const res = fn("books");

It seems like a bug in TypeScript that it doesn't allow such things, that's why you need the as any . See here for an interactive example.