How to define a function's argument type as one of the keys (or properties) of an interface

Question

Given any interface, is there a way to say that a variable's type is one of the keys in that interface?

Suppose you have this interface:

interface IExample {
  a: string;
  b: {
    b1: string;
    b2: string | number | boolean;
  };
}

And you have a function like:

const foo = (arg) => {
  //function's logic 
}

Now I want to type arg as being b from IExample , something like:

const foo = (arg: IExample.b): void => {  // this generates an error
  //function's logic 
}

---

To be clear, the function's argument should be:

{
  b1: string;
  b2: string | number | boolean;
}

and I didn't want to have to write another interface just for that.

I couldn't find a way by myself, neither figure it out by reading the typescript docs. This is my last hope.

Answer 1

You use [] to index types. Some documentation on this is here.

function foo(arg: IExample['b']): void {
  //function's logic 
}

foo({ b1: 'abc', b2: true })

See Playground