Question

由于尚未实现模块的条件导入，因此如何扩展或覆盖同一类：

// I have a main class
// main.js
export default class Main {}

// Main class is used by myClass
class myClass {
  constructor(){
    new Main()
  }
}

// I have an extended main class
// main-extended.js
export default class MainExtended extends Main {}

// the MainExtended is now used by myClass
export class myClass {
  constructor(){
    new MainExtended()
  }
}


// third-class.js
// now I need to do this with any class I can find, 
// preferably with the extended version of both, if both defined
export default class ThirdClass {
  constructor(...args){
    return new myClass(...args)    
  }
}


// destinations

// index.js
import Main, {myClass} from main.js
import ThirdClass from third-class.js


// index-extended.js
import MainExtended, {myClass} from main-extended.js
import ThirdClass from third-class.js

问题是，由于某种原因，index.js编译后还包含了myClass要求MainExtended，因此我决定只在ThirdClass中包含index-extended ，作为MainExtended的奖励功能。

但是我仍然想知道是否还有另一种方法。

Answer 1

像VLAZ在评论中说的那样，我不确定您要问的是什么，但您能做到100％这样吗？

解决方案1 

class MyClass {
  constructor(_main){
    myMain = _main;
  }
}

const regularMain = new MyClass(new Main());
const extendedMain = new MyClass(new MainExtended());

ES6 / ES7扩展具有相同名称的类

1 个答案: