Typescript 2.8：从一种类型中删除另一种类型的属性

Question

在更新日志2.8中，他们有条件类型的示例：

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"

除了删除对象的属性之外，我想这样做。我怎样才能实现以下目标：

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，利用之前的Diff类型（顺便说一下，它与现在属于标准库的Exclude类型相同），您可以写：

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