Javascript 将类的私有成员设置为构造函数参数

发布于05月24日

class Foo {
  #one
  #two
  #three
  #four
  #five
  #six
  #seven
  #eight
  #nine
  #ten
  #eleven
  #twelve
  #thirteen
  #fourteen
  #fifteen
  #sixteen

  constructor(
    one,
    two,
    three,
    four,
    five,
    six,
    seven,
    eight,
    nine,
    ten,
    eleven,
    twelve,
    thirteen,
    fourteen,
    fifteen,
    sixteen
  ) {
    this.#one = one;
    this.#two = two;
    this.#three = three;
    this.#four = four;
    this.#five = five;
    this.#six = six;
    this.#seven = seven;
    this.#eight = eight;
    this.#nine = nine;
    this.#ten = ten;
    this.#eleven = eleven;
    this.#twelve = twelve;
    this.#thirteen = thirteen;
    this.#fourteen = fourteen;
    this.#fifteen = fifteen;
    this.#sixteen = sixteen;
  }
}

这种(反？)模式的解决方案是什么？

推荐答案

对于任何希望使用您的构造函数的人来说，拥有16个参数都不是一件有趣的事情.您在注释中提出的配置对象概念要有趣得多，当然，当您将其与拥有一个具有所有这些属性的Object类型的私有属性的 idea 结合在一起时.然后，您可以使用Object.assign将其更新为用户的首选项:

class Foo {
  #options = {
    one: 1,
    two: 2,
    three: 3,
    four: 4
  }
  constructor(options = {}) {
    Object.assign(this.#options, options);
    console.log(this.#options);
  }
}

let foo = new Foo({three: 3000});