Question

我有以下课程结构......

export abstract class PersonBase {
    public toJSON(): string {
        let obj = Object.assign(this);
        let keys = Object.keys(this.constructor.prototype);
        obj.toJSON = undefined;
        return JSON.stringify(obj, keys);
    }
}

export class Person extends PersonBase {

    private readonly _firstName: string;
    private readonly _lastName: string;

    public constructor(firstName: string, lastName: string) {
        this._firstName = firstName;
        this._lastName = lastName;
    }

    public get first_name(): string {
        return this._firstName;
    }

    public get last_name(): string {
        return this._lastName;
    }
}

export class DetailPerson extends Person {

    private _address: string;

    public constructor(firstName: string, lastName: string) {
        super(firstName, lastName);
    }

    public get address(): string {
        return this._address;
    }

    public set address(addy: string) {
        this._address = addy;
    }
}

我试图让jSON（）从完整的对象层次结构中输出所有的getter（不包括私有属性）......

所以如果我有一个DetailPerson实例并且我调用.toJSON（）我想看到以下输出......

{ “地址”：“一些地址”， “first_name”：“我的名字”， “last_name”：“我的姓” }

我使用了本文中的一个解决方案，但它没有解决我的特定用例......我没有在输出中获得所有的getter。

Using JSON.stringify in conjunction with TypeScript getter/setter

我需要在此更改以获得我想要的结果？

Answer 1

您提供的链接使用Object.keys，这会在原型上留下属性。

您可以使用for...in代替Object.keys：

public toJSON(): string {
    let obj: any = {};

    for (let key in this) {
        if (key[0] !== '_') {
            obj[key] = this[key];
        }
    }

    return JSON.stringify(obj);
}

编辑：这是我尝试以递归方式返回getter，而不假设非getter以下划线开头。我确定我错过了遗漏（循环引用，某些类型的问题），但这是一个好的开始：

abstract class PersonBase {
  public toJSON(): string {
    return JSON.stringify(this._onlyGetters(this));
  }

  private _onlyGetters(obj: any): any {
    // Gotchas: types for which typeof returns "object"
    if (obj === null || obj instanceof Array || obj instanceof Date) {
      return obj;
    }

    let onlyGetters: any = {};

    // Iterate over each property for this object and its prototypes. We'll get each
    // property only once regardless of how many times it exists on parent prototypes.
    for (let key in obj) {
      let proto = obj;

      // Check getOwnPropertyDescriptor to see if the property is a getter. It will only
      // return the descriptor for properties on this object (not prototypes), so we have
      // to walk the prototype chain.
      while (proto) {
        let descriptor = Object.getOwnPropertyDescriptor(proto, key);

        if (descriptor && descriptor.get) {
          // Access the getter on the original object (not proto), because while the getter
          // may be defined on proto, we want the property it gets to be the one from the
          // lowest level
          let val = obj[key];

          if (typeof val === 'object') {
            onlyGetters[key] = this._onlyGetters(val);
          } else {
            onlyGetters[key] = val;
          }

          proto = null;
        } else {
          proto = Object.getPrototypeOf(proto);
        }
      }
    }

    return onlyGetters;
  }
}

JSON.stringify仅序列化TypeScript getter

1 个答案: