在JS中的一组构造函数中反结构参数

Question

我如何才能1)将原型道具从父构造函数传递给子构造函数，2)使用destructure来组织所有构造函数的参数，而不引起类型错误？

这就是我想修正的代码。

function Participant ({gender, age, tastes, education}){
  this.gender = gender;
  this.age = age; 
  this.tastes = tastes || [];
  this.education = education || 0;
}

function Payer(gender, age, tastes, education, income, job){
  this.base = Participant;
  this.base(gender, age, tastes, education); 
  this.income = income;
  this.job = job || 'merchant';
}
Payer.prototype = new Participant;

function NotPayer({gender, age, tastes, role}){
  this.role = role || 'kid';
  this.base = Participant;
  this.job = 'Not Valid';
  this.base(gender, age, tastes);
}
NotPayer.prototype = new Payer;

const kid1 = new NotPayer({role: 'nephew', age: 6, gender: 'male', tastes: ['ps4', 'golf']});
kid1.role;// 'nephew'
const kid2 = new Participant({gender: 'female'});
kid2.gender; // Type Error: Cannot destructure property 'gender' of 'undefined'

我试图修复的是这样的：

function Participant ({gender, age, tastes, education}){
  this.gender = gender;
  this.age = age; 
  this.tastes = tastes || [];
  this.education = education || 0;
}

function Payer(gender, age, tastes, education, income, job){
  this.base = Participant;
  this.base(gender, age, tastes, education);
  this.income = income;
  this.job = job || 'merchant';
}
//Payer.prototype = new Participant;

function NotPayer({gender, age, tastes, role}){
  this.role = role || 'kid';
  this.base = Participant;
  this.job = 'Not Valid';
  this.base(gender, age, tastes);
}
//NotPayer.prototype = new Payer;

const kid1 = new NotPayer({role: 'nephew', age: 6, gender: 'male', tastes: ['ps4', 'golf']});
console.log(kid1.role); // nephew
const kid2 = new Participant({gender: 'female'});
console.log(kid2.gender); // female

这只是因为我在表达式中添加了注释，将父构造函数分配给子构造函数（结果是，例如： kid1.gender 成为 undefined)

注意到 这段代码可以同时使用父原型继承和参数析构，但只有在下面的情况下（在最后一个构造子函数中使用析构时）。

function Participant (gender, age, tastes, education){
  this.gender = gender;
  this.age = age; 
  this.tastes = tastes || [];
  this.education = education || 0;
}
function NotPayer({gender, age, tastes, role}){
  this.role = role || 'kid';
  this.base = Participant;
  this.job = 'Not Valid';
  this.base(gender, age, tastes);//check education
}
NotPayer.prototype = new Participant;

const kid1 = new NotPayer({role: 'nephew', age: 6, gender: 'male', tastes: ['ps4', 'golf']});
console.log(kid1.gender); // male

console.log(NotPayer.prototype); // Participant{...}

谢谢！！很抱歉这个问题的范围。

Answer 1

试试这个

class Participant {
  constructor(gender, age, tastes, education) {
    this.gender = gender;
    this.age = age;
    this.tastes = tastes || [];
    this.education = education || 0;
  }
}

class NotPayer extends Participant {
  constructor({gender, age, tastes, role}) {
    super(gender, age, tastes)
    this.role = role || 'kid';
    this.job = 'Not Valid';
  }
}

const kid1 = new NotPayer({role: 'nephew', age: 6, gender: 'male', tastes: ['ps4', 'golf']});
console.log(kid1.gender)

console.log(NotPayer.prototype)