Why does typescript give a type error on type unions?

Question

Whenever I have to work with objects that have a combined union type, typescript complains about the properties that I try to access and I don't get autocompletion either. For example this:

interface A {
  id: string;
  value: number;
}

interface B {
  result: string;
}

export type Types = A | B;

function test(obj: Types) {
  obj.result;  // want to work with obj as though it implements interface B
}

I get errors when I access result, id, and value from typescript:

Property 'result' does not exist on type 'Types'.
  Property 'result' does not exist on type 'A'

Is there any way that I can narrow down the interface type so that I would get a better IDE experience?

You need a [*type guard*](https://www.typescriptlang.org/docs/handbook/advanced-types.html#type-guards-and-differentiating-types) in one way or another. — Yoshi, Jul 15 '20 at 08:11
Because `result` only exists on `B`, if your `obj` is a `A` , then accessing `result` on it will return `undefined` — Trash Can, Jul 15 '20 at 08:11

score 5 · Accepted Answer · answered Jul 15 '20 at 08:14

interface A {
  type:'A';
  id: string;
  value: number;
}

interface B {
  type:'B';
  result: string;
}

export type Types = A | B;

function test(obj: Types) {
  if(obj.type==='B'){
    obj.result;
  }
}

You need a common field to teach TS how to recognize type A or B.

https://www.typescriptlang.org/docs/handbook/advanced-types.html#discriminated-unions

Stark Jeon · Answer 2 · 2020-07-15T08:27:32.937

If you want that action you can use & operator like

type Types = A & B

you can merge A interface and B interface. It's naming Intersection Types

And I can write some info about this. Can you wait some minutes?

Reference

And now I know that what I got it was wrong. So if you want to use type gaurd it's work for you.

enum ALPHA_TYPE {
   A = "A",
   B = "B"
}
interface A {
   ...
}
interface B {
   ...
}
export type Types<T extends ALPHA_TYPE> = T extends A ? A : B

function test<T>(obj: Types<T>) {
  obj.result;  // want to work with obj as though it implements interface B
}

and you can use like

text<ALPHA_TYPE.A>()

Sure, would love an explanation. Have been searching for an answer for very long — Loupi, Jul 15 '20 at 08:13

score 0 · Answer 3 · answered Nov 30 '21 at 21:30

interface A {
  id: string;
  value: number;
}

interface B {
  result: string;
}

 type Types = A | B;

function test(obj: Types) {
  const objB = obj as B; // Introduce a new variable and cast it as B
  objB.result; // Happy!!!
}

Just introduce a new variable and cast it to B, then you get to call .result on the object.

Why does typescript give a type error on type unions?

3 Answers3

Reference