Kevin Powell - Community•10mo ago

Conditional Type

Is it possible to have a type which resolves between two types with a conditional whether a field on an object exists or not.

Type A = {a: string, b: string}
Type B = {c: string, b: number}

Type C = A | B

Type A = {a: string, b: string}
Type B = {c: string, b: number}

Type C = A | B

I want to achieve something like this

const obj: C;
obj.a ? obj.b : obj.b + 1

const obj: C;
obj.a ? obj.b : obj.b + 1

I know the example can be resolved with another way but my use case is more complex. I just want the type to be resolved like that. How do you go about this?

2 Replies

👾 | Lofty | 👾•10mo ago

the only way if extending a type would be like

type Matches<T> = T extends A ? B : T;

type Matches<T> = T extends A ? B : T;

👾 | Lofty | 👾•10mo ago

https://www.typescriptlang.org/docs/handbook/2/conditional-types.html#conditional-type-constraints

Documentation - Conditional Types

Create types which act like if statements in the type system.

Gaming

Programming

Conditional Type