Why does typescript give a type error on type unions?
Asked Answered
S

3

6

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?

Serration answered 15/7, 2020 at 8:8 Comment(2)
You need a type guard in one way or another.Beeler
Because result only exists on B, if your obj is a A , then accessing result on it will return undefinedCand
F
6
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

Flawy answered 15/7, 2020 at 8:14 Comment(0)
F
3

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

  1. About & Intersection Types
  2. About Union Tpyes

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>()
Fineable answered 15/7, 2020 at 8:11 Comment(1)
Sure, would love an explanation. Have been searching for an answer for very longSerration
S
0
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. enter image description here

Spit answered 30/11, 2021 at 21:30 Comment(0)

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