TypeScript conditional types for props

Question

I have a component that can be in two different states, editing mode and viewing mode. To accommodate for that I have a union type to make this work, but I still get TypeScript complaining in the args of my component:

type View = {
    isEditing: false;
    viewHandler: () => void;
    someOtherProp: any // this is unique to view mode
};

type Edit = {
    isEditing: true;
    editHandler: (id: string) => void;
};

export type MyProps = View | Edit;

It is then complaining here:

const Item: React.FC<MyProps> = ({
    isEditing = false,
    editHandler,
    ~~~~~~~~~~~
}) => {

Property 'editHandler' does not exist on type 'PropsWithChildren<MyProps>'.

What am I missing?

Answer 1

So the solution I used now is adding it and typing it as never:

type View = {
    isEditing: false;
    viewHandler: () => void;
    editHandler?: never;
};

type Edit = {
    isEditing: true;
    editHandler: (id: string) => void;
    viewHandler?: never;
};

export type MyProps = View | Edit;

And at the point where the event handler gets invoked or passed down, I use a ! to assure TypeScript that this cannot be undefined. Kudos to Nadia who pointed me in the right direction in a comment above!