To make a union type from a type alias or interface properties, you can use the indexed access type syntax and then write the property names to make the union type separated by the union operator symbol (|) in TypeScript.
TL;DR
// a simple interface
interface Car {
name: string;
yearMade: number;
}
// make a union type from the `name` and
// `yearMade` properties from the `Car` interface
type NameOrYear = Car["name" | "yearMade"]; // NameOrYear -> string | number
For example, let's say we have an interface called Car having a property called name with the type of string and another property called yearMade having the type of number like this,
// a simple interface
interface Car {
name: string;
yearMade: number;
}
Suppose we want to make a union type from the name and yearMade properties from the Car interface.
To do that we can make use of the indexed access type syntax and then write out the property names to make the union type separated by the union operator symbol (|).
It can be done leik this,
// a simple interface
interface Car {
name: string;
yearMade: number;
}
// make a union type from the `name` and
// `yearMade` properties from the `Car` interface
type NameOrYear = Car["name" | "yearMade"]; // NameOrYear -> string | number
Now if you hover over the NameOrYear type, you can see that it is a union type composed of string and number type which are the same types of the name and yearMade properties from the Car interface.
We have successfully made a union type from interface properties in TypeScript. Yay 🥳!
This method is the same for any type alias type in TypeScript.
See the above code live in codesandbox.
That's all 😃!