typescript - 受保护属性的通用 getter 和 setter
问题描述
考虑这个 TypeScript 类,它通过将实现拆分为超类和子类来提供类型检查的通用 getter 和 setter:
class ICompetence {
protected _id : number | undefined;
protected name: string | undefined;
}
export default class Competence extends ICompetence {
set<K extends keyof ICompetence>(key: K, value: ICompetence[K]): void {
this[key] = value;
}
get<K extends keyof ICompetence>(key: K) : ICompetence[K] {
return this[key];
}
constructor() {
super();
}
}
该类的使用方式如下:
let competence = new Competence():
competence.set("name", "name value text");
但是,然后我收到此错误:
error TS2345: Argument of type '"name"' is not assignable to parameter of type
可以通过从接口类中删除受保护的注释来克服该错误:
class ICompetence {
_id : number | undefined;
name: string | undefined;
}
TypeScript 是否提供任何功能来避免在保护属性时发生这种错误?
解决方案
由于protected
仅在设计时生成错误并且不会在运行时阻止访问,因此您可能会以另一种方式获得类似的行为,即从导出的类型中隐藏相关属性,如下所示:
首先,公开父类的属性:
class ICompetence {
_id: number | undefined;
name: string | undefined;
}
然后正常扩展类,但不要导出它,并将其重命名为_Competence
(“真实”Competence
将在稍后出现)
class _Competence extends ICompetence {
otherProperty: string = "hmm"; // demonstrate you can add other things
set<K extends keyof ICompetence>(key: K, value: ICompetence[K]): void {
this[key] = value;
}
get<K extends keyof ICompetence>(key: K): ICompetence[K] {
return this[key];
}
constructor() {
super();
}
}
现在我们准备Competence
。首先,我们定义类型函数Omit
,它从类型中删除指定的键:
type Omit<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>
然后我们导出一个名为的接口Competence
和一个名为的值Competence
,它们派生自_Competence
但不暴露其属性ICompetence
。当然,它们仍然存在,但类型系统不会暴露它们:
export interface Competence extends Omit<_Competence, keyof ICompetence> { }
export const Competence = _Competence as new () => Competence;
现在在您的消费者代码中,您应该能够执行以下操作:
const competence = new Competence();
competence.name // error at compile time
competence.otherProperty // okay
const name = competence.get("name") // string or undefined
它有效。希望有帮助。祝你好运!