TypeScript: remove index signature using mapped types

Question

Given an interface (from an existing .d.ts file that can\'t be changed):

interface Foo {
  [key: string]: any;
  bar(): void;
}

Is there a

Answer 1

There is a way, requiring TypeScript 2.8's Conditional Types.

It is based on the fact that 'a' extends string but string doesn't extends 'a'

interface Foo {
  [key: string]: any;
  bar(): void;
}

type KnownKeys = {
  [K in keyof T]: string extends K ? never : number extends K ? never : K
} extends { [_ in keyof T]: infer U } ? U : never;


type FooWithOnlyBar = Pick>;

You can make a generic out of that:

// Generic !!!
type RequiredOnly> = Pick>;

type FooWithOnlyBar = RequiredOnly;

For an explanation of why exactly KnownKeys works, see the following answer:

https://stackoverflow.com/a/51955852/2115619