In the 2.8 的变更日志,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"
我想做到这一点,除了删除对象的属性。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; }
好吧,利用你的Diff
之前的类型(顺便说一句,这与Exclude
type 现在是标准库的一部分),你可以写:
type ObjectDiff<T, U> = Pick<T, Diff<keyof T, keyof U>>;
type AWithoutStuff = ObjectDiff<A, Stuff>; // inferred as { one: string; two: number; }
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