typescript - Declare a type that is a union of the types of the properties of an interface - Stack Overflow

I have an interface:

{
   a: string,
   b: number
}

I want to make a type that can, based on the above interface, be either a string or a number. So, a type that 'does the same thing' as the following:

type MyType = string | number

But created based from the interface, like type MyType = {any type from the properties of the interface}

I have an interface:

{
   a: string,
   b: number
}

I want to make a type that can, based on the above interface, be either a string or a number. So, a type that 'does the same thing' as the following:

type MyType = string | number

But created based from the interface, like type MyType = {any type from the properties of the interface}

• typescript

Share Improve this question Follow edited Jan 29 at 9:21 jonrsharpe 122k3030 gold badges268268 silver badges476476 bronze badges asked Jan 29 at 9:11 Dan.Dan. 72711 gold badge88 silver badges2323 bronze badges 1

• 1 Thing[keyof Thing]? – jonrsharpe Commented Jan 29 at 9:21

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1 Answer 1

Sorted by: Reset to default 1

You can use it like this:

interface MyInterface {
   a: string;
   b: number;
}

type MyType = MyInterface[keyof MyInterface];

const test1: MyType = "abc";
const test2: MyType = 55;