Question

我正在试图弄清楚这里发生了什么，因为在初始构建之后，Parent / Super类没有数据。

// imports / server / a-and-b.js

class A {
  constructor(id) {
    // make MongoDB call and store inside this variable
    // ...
    this._LocalVariable = FieldFromMongo;
    console.log(`this._LocalVariable: ${this._LocalVariable}`); // => This has a good value, ie: 'Test'
  }
  get LocalVar() {
    console.log(`this._LocalVariable: ${this._LocalVariable}`); // => This has a undefined value when called from child class
    return this._LocalVariable;
  }
}

export class B extends A {
  constructor(id) {
    super(id);
    this.TEST = 'THIS IS A TEST';
  }
  get THE_Variable() {
    console.log(`super.LocalVar: ${super.LocalVar}`); // => This has a undefined value when called
    return super.LocalVar;
  }
  get GETTHEVAR() {
    return this.TEST; // => This returns 'THIS IS A TEST'
  }
}

// imports / server / factory.js

import { B } from 'imports/server/a-and-b.js';

class Factory {
  constructor() {
    this._factory = new Map();
  }
  BuildInstances(id, cls) {
    let instance = this._factory.get(cls);

    if (!instance) {
      if (cls === 'B') {
        instance = new B(id);
        this._factory.set(cls, instance);
        return instance;
      }
    }
    else {
      return instance;
    }
  }
}
export let OptsFactory = new Factory();

// imports / server / test.js

import { OptsFactory } from 'imports/server/factory.js'

const B = OptsFactory.BuildInstances(id, 'B');

const THE_Variable = B.THE_Variable; // => always undefined

const TEST = B.GETTHEVAR; // => Always returns 'THIS IS A TEST'

为什么A班没有保持状态？

Answer 1

这是我发现的：

class A {
  constructor(id) {
    // make MongoDB call and store inside this variable
    // ...
    this._LocalVariable = FieldFromMongo;
  }
  get LocalVar() {
    return this._LocalVariable;
  }
  GetThatLocalVar() {
    return this._LocalVariable;
  }
}

export class B extends A {
  constructor(id) {
    super(id);
  }
  get Style1() {
    // Reference to Parent get function
    return super.LocalVar; // => This has a undefined value when called
  }
  get Style2() {
    // Reference to Parent property
    return super._LocalVariable; // => This has a undefined value when called
  }
  get Style3() {
    // Reference to local Property that is declared in Parent
    return this._LocalVariable; // => This works
  }
  get Style4() {
    // Reference to Parent without the getter
    return super.GetThatLocalVar(); // => This works
  }
  get GETTHEVAR() {
    return this.TEST; // => This returns 'THIS IS A TEST'
  }
}

所以基本上可行的是Style3 Style 4的工作。

ES6：超级级别没有状态

1 个答案: