How to make that one property of an object parameter required and the rest optional in TypeScript?

Question

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers. question from:https://stackoverflow.com/questions/65851607/how-to-make-that-one-property-of-an-object-parameter-required-and-the-rest-optio

Answer 1

Use a type or an interface, then declare optional properties (using ?).

Using a type:

type Sweets = {
  candy?: string;
  chocolate: string;
  lollipop?: string;
}

Using an interface:

interface Sweets {
  candy?: string;
  chocolate: string;
  lollipop?: string;
}

And then in your script:

const sweets: Sweets = {
  candy: 'Skittles',
  chocolate: 'Milka',
  lollipop: 'ChupaChups',
};

const getChocolate = ({ chocolate }: Sweets): boolean => chocolate.includes('something');

Update to define type on function instead of on argument

You can use a generic type on your function to dynamically-type the function's argument. I believe something like this should do what you want:

/*

'T' is a generic type, you can set the type
when you _call_ the function,
not when you define it
extending Record<'chocolate', string> means that
although we don't know the argument type yet,
we know it should at least have a chocolate: string property

*/
const getChocolate = <T extends Record<'chocolate', string>>(
  { chocolate }: T,
): boolean => chocolate.includes('something');

// the type for arg `{ chocolate }` will be set when the function is _called_

// argument for `{ chocolate }` is now of type 'Sweets'
getChocolate<Sweets>(mySweets);