由于模块的条件导入尚未实现,我如何扩展或重写同一个类:
// 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
编译还包括myClass
调用MainExtended
,所以我决定只包括index-extend
中的 ThirdClass
,作为 MainExtended
的额外功能。
但我还是想知道是否还有其他方法。
最佳答案
就像 VLAZ 在评论中所说的那样,我不能 100% 确定你在问什么,但是你能做这样的事情吗?
解决方案1
class MyClass {
constructor(_main){
myMain = _main;
}
}
const regularMain = new MyClass(new Main());
const extendedMain = new MyClass(new MainExtended());
关于javascript - ES6/ES7 扩展同名类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59688253/