Can I use a condition to define a type in an interface?

Question

I have this interface for an item:

interface IItem {
  id: string;
  name: string;
  description: string;
  note: string | undefined;
  type: string;
  sellable: boolean;
  buyable: boolean;
  usable: boolean;
  price: number | undefined; // <--- only undefined when both buyable and sellable are false
  usage: string;
}

Because I define the items, I know that price will always be a number unless sellable and buyable is false. But as of right now, TypeScript doesn't know that. Which causes me to have to add tautologic typeguards: if (buyable) will still require another if statement: if (price), even though I know that will always be a number because of the previous if statement.

Pseudo example:

  sellable: bool
  buyable: bool
  price: if !sellable & !buyable: undefined else number

Answer 1

You can accomplish this using some additional types and type union.

Start defining your types

type A1 = {
  sellable: false;
  buyable: false;
  price: undefined;
};

type A2 = {
  sellable: boolean;
  buyable: true;
  price: number;
};

type A3 = {
  sellable: true;
  buyable: boolean;
  price: number;
};

Then, create a union type using all types defined above:

type A = A1 | A2 | A3;

Finally, you can define your type IItem as

type IItem = A & {
  id: string;
  name: string;
  description: string;
  note: string | undefined;
  type: string;
  usable: boolean;
  usage: string;
}