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Typescript 2.8:从一种类型中删除另一种类型中的属性

conditional-types typescript2.8 typescript

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Rico Kahler · 技术社区 · 8 年前

在 changelog of 2.8 ,条件类型的示例如下:

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"

我只想删除对象的属性。如何实现以下目标:

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; }

1 回复 | 直到 7 年前

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6

CRice 8 年前

好吧,利用你的 Diff 键入早期版本(顺便说一下,与 Exclude 类型(现在是标准库的一部分),您可以编写:

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