Typescript 2.8：從一種類型中刪除另一種類型的屬性

Question

在更新日志2.8中 ，他們有條件類型的這個例子：

type Diff<T, U> = T extends U ? never : T;  // Remove types from T that are assignable to U
type T30 = Diff<"a" | "b" | "c" | "d", "a" | "c" | "f">;  // "b" | "d"

除了刪除對象的屬性之外，我想這樣做。 我怎樣才能實現以下目標：

type ObjectDiff<T, U> = /* ...pls help... */;

type A = { one: string; two: number; three: Date; };
type Stuff = { three: Date; };

type AWithoutStuff = ObjectDiff<A, Stuff>; // { one: string; two: number; }

Answer 1

好吧，利用前面的Diff類型（順便說一下，它與現在屬於標准庫的Exclude類型相同），你可以寫：

type ObjectDiff<T, U> = Pick<T, Diff<keyof T, keyof U>>;
type AWithoutStuff = ObjectDiff<A, Stuff>; // inferred as { one: string; two: number; }