Define a new type in Typescript that is equal to another type, but with the added flexibility of having optional

I have 2 categories:

category Main = {
  x: boolean;
  y: number;
  z: string
}

category MainOptions = {
  x?: boolean;
  y?: number;
  z?: string;
}

In this scenario, MainOptions is designed to include some, none, or all of the attributes that belong to Main based on the specific situation, but never to contain an attribute that does not exist in Main.

I am aiming to connect them systematically so that as Main evolves and gains new attributes, I only need to modify the Main type.

What is the best way to accomplish this?

Answer №1

If you want to use a type in TypeScript and make all properties optional, you can employ the Utility type known as Partial. Check out the Official Documentation for more information.

Here is an example taken from the TypeScript Documentation:

interface Task {
  name: string;
  description: string;
}
 
function updateTask(task: Task, fieldsToUpdate: Partial<Task>) {
  return { ...task, ...fieldsToUpdate };
}
 
const task1 = {
  name: "clean room",
  description: "tidy up and dust shelves",
};
 
const task2 = updateTask(task1, {
  description: "vacuum floor",
});

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