是否可以使用打字稿映射的类型来创建接口的非功能属性类型? [英] Is it possible to use typescript mapped types to make a type of non-function properties of an interface?

问题描述

所以我在看Typescript的映射类型.是否可以创建一个包装另一个类型的接口，以从原始类型中删除功能?例如:

so I was looking at Typescript's mapped types. Would it be possible to create an interface that wraps another type that removes functions from the original type? For example:

interface Person{
name: string,
age: number,
speak(): void,
}

type Data<T> = ?

const dataPerson: Data<Person> ={
name: "John",
age: 20
//Speak removed because it is a function
};

谢谢！

推荐答案

这是来自打字稿文档(

This is from the typescript documentation (https://www.typescriptlang.org/docs/handbook/advanced-types.html#conditional-types) and works:

type NonFunctionPropertyNames<T> = { [K in keyof T]: T[K] extends Function ? never : K }[keyof T];
type Data<T> = Pick<T, NonFunctionPropertyNames<T>>;

谢谢大家！

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