What is the method for creating an object type that necessitates a key determined by a variable?

Is it feasible to create a custom type in TypeScript that can check if a given key exists within the Type definition? I am specifically using noUncheckedIndexAccess: true configuration.

interface one {
  foo: string;
  bar: string;
}

interface two {
  bar: string;
  foobar: string;
}

interface three {
  foobar: string;
  foo: string;
}

interface SomeType<T> {
  ..?..
}

function fn<T>(object: SomeType<T>, key: T) {
  object[key] // is defined
  //...
}

fn<"foo">(arg); // Typeof arg must be (one | three)
fn<"bar">(arg); // Typeof arg must be (one | two)

Answer №1

If you have a combination of all the permitted types you wish to utilize:

type Allowed = one | two | three;

as well as a method to retrieve all the keys of the permissible types:

type KeysOfUnion<U> = U extends U ? keyof U : never;
// KeysOfUnion<Allowed>

then it is possible to define your fn in this manner:

function fn<T extends KeysOfUnion<Allowed>>(key: T, object: Extract<Allowed, Record<T, any>>) {
  const thing = object[key];
  //    ^?
}

where the type of object is a subset of the allowed types where the elements possess the specified key.

The screenshot provided displays the inferred type of object based on the given key:

https://i.stack.imgur.com/F9fGe.png

Playground

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