TS: Construct type from optional properties of interface-CodePudding

Supposing we have a similar interface:

interface Interface {
  foo: string;
  bar: number;
  baz?: boolean;
}

How can we construct a type with only its optional properties?

With the interface above, I'd be looking to something like:

interface PartialInterface {
  baz: boolean;
}

A solution would be of course to use either Omit or Pick; That would work on small interfaces, but it wouldn't really be suitable for bigger scenarios.

Any suggestion is highly appreciated, thank you in advance!

CodePudding user response：

We'll combine the removal of the ? modifier using the - prefix and filtering properties with the as keyword in mapped types:

type PickOptionals<T> = {
    // `Interface` below extends
    //   `Record<"foo", string>` and
    //   `Record<"bar", number>` but not
    // - `Record<"baz", boolean>`.
    // Since the right side of `as` evaluates to `never` when
    // `K` is "foo" or "bar", those properties are dropped entirely.
    [K in keyof T as T extends Record<K, T[K]> ? never : K]-?: T[K];
}

interface Interface {
  foo: string;
  bar: number;
  baz?: boolean;
}

type PartialInterface = PickOptionals<Interface>
// = { baz: boolean }

Here's a playground link for good measure