Conditional boolean type in typescript with interface

Question

How would I write conditional boolean type for this?

ie

export interface Props {
  editable: boolean;
  position: { editable: true } ? Sheets.Matrix : never;
  value: string;
  classNames?: string;
  textType?: "bold" | "editing" | "normal";
  onChange?: (val: Sheets.SaveData) => void;
}

I use props like this

const SheetsCell: React.FC<MainProps> = ({
  editable,
  value,
  textType = "normal",
  classNames = "",
  position,
  ...

Here if something is editable, I want the type of position to be Sheets.Matrix else, position isn't required.

Answer 1

One way to do it is using an interface and the intersection operator( & ).

You can use an intersection type , which combines multiple types into one.

interface Props {
  value: string;
  classNames?: string;
  textType?: "bold" | "editing" | "normal";
  onChange?: (val: number) => void;
}

type MainProps = Props & {
  editable: true;
  position: number; 
} | Props & {
  editable: false;
  position?: number; 
}; 

The above type MainProps forces editable to be true in one case and false in another. If editable is true , then position property exists and if it is false, position property exists but it is not required.

unions is used above, to generate a type which can be one of the two values. MainProps can be one of the two types (both being intersection with Props type).

Here is the playground link

Answer 2

You could use type aliases instead of an interface eg

type Props = { 
  editable: true;
  position: Sheets.Matrix;
} | { 
  editable: false;
};

const testFalse: Props = { // position not required
  editable: false
}
    
const testTrue: Props = { // position mandatory
  editable: true,
  position: MatrixEnum
}