如何将 object 的部分属性键提取为联合类型？

Question

In Typescript, I'd like to do the following:

Typescript playground

type MY_OBJECT = {
  fooA: string,
  fooB: string,
  fooD: string,
  fooE: string,
  fooF: string
}

type SOME_PROPS = Extract<keyof MY_OBJECT, "fooA" | "fooB" | "fooC">

I'd like Typescript to "know" that there's no "fooC" available on keyof MY_OBJECT and warn me about that.

In other words, I'd like to extract part of the properties of an object and a union type and I'd like to keep it all strongly typed.

For example, if I ever change fooA to fooAAA on MY_OBJECT , I'd like to know immediately that I need to fix something on that Extract<> type, since fooA would no longer exist.

Is it possible?

My use case: I have a Redux action that is supposed to update some of the properties of a specific object. So I'll be using that union type (composed by the selected property names) to type my action and make sure that the action receive only the allowed names.

type MY_OBJECT = {
  fooA: string,
  fooB: string,
  fooD: string,
  fooE: string,
  fooF: string
}

type SOME_PROPS = Extract<keyof MY_OBJECT, "fooA" | "fooB">
type UPDATE_SOME_PROPS  = PayloadAction<{ name: SOME_PROPS , value: SOME_VALUE }>

Answer 1

Here is what I've ended up doing:

Typescript playground

I'm using keyof Pick<> . This way you get intellisense for the extracted properties.

type MY_OBJECT = {
  fooA: string,
  fooB: string,
  fooD: string,
  fooE: string,
  fooF: string
}

type SOME_PROPS = keyof Pick<MY_OBJECT, "fooA" | "fooB">
type SOME_PROPS_TEST = keyof Pick<MY_OBJECT, "fooA" | "fooB" | "fooC">

type SOME_PROPS = keyof Pick<MY_OBJECT, "fooA" | "fooB">                   // WORKS
type SOME_PROPS_TEST = keyof Pick<MY_OBJECT, "fooA" | "fooB" | "fooC">     // TRIGGERS ERROR