Question

有什么办法可以在打字稿中得到这个结果吗？有没有办法直接从类名称而不是从对象获取父类的属性列表我的最终目标是获得这个结果

export class A {
    public a: string;
    public b: string;
}
export class B extends A {
    public missingProperty: any;
    public constructor(res: any){
      super();
      this.missingProperty = {};
      this.setMissingProperty(res);
    }
    private setMissingProperty(res: any){
       Object.keys(res).foreach(element => {
           // if property name match i.e a is present in res as well as class B then assign to the same 
           // property name 
           // else assign to missingProperty
       }
    }
}

我想要什么

// test
const temp = {a:'testa', b: 'testb', 0: 23, 1: 34, 2:75}
const b = new B(temp);
// b.b gives testb
// b.a gives testa
// b.missingProperty gives {0: 23, 1: 34, 2:75}

Answer 1

TypeScript编译为JavaScript运行时。显然，在类上声明字段不会初始化它：

console.log(new A()) // prints: { __proto__: {...}}
                     // note how it does not print {a: undefined, b: undefined}
                     // so right now you have no idea what properties A has!
console.log(new B()) // prints: { missingProperty: {}, __proto__: { ... } }
                     // you have no idea what properties A has either!

如果您想在运行时知道哪些属性属于对象，那么您的问题本质上不是TypeScript问题-它是 Javascript 问题！

现在，没有办法知道运行时是否存在A中的属性。相反，您可以做的是首先使用A的构造函数初始化这些已知的属性，因此setMissingProperty将会知道B已经具有这些A继承的属性。然后它可以遍历并查找B没有的属性。

class A {
    public a: string;
    public b: string;
    constructor(options) {
      this.a = options.a;
      this.b = options.b;
    }
}
class B extends A {
    public missingProperty: any;
    public constructor(res: any){
      super(res); // B.a, B.b is initialized here
      this.missingProperty = {};
      this.setMissingProperty(res);
    }
    private setMissingProperty(res: any){
      Object.keys(res).forEach(element => {
        if (!Object.prototype.hasOwnProperty.call(this, element)) {
          this.missingProperty[element] = res[element]
        }
      })
    }
}

获取父类属性列表

1 个答案: