在 typescript 中的基础 class 的子类中实现的调用方法

Question

//BaseClass.ts

class Base {
  connstructor() {}

  callSubClassMethod = () => {
    return this.subMethod();
  }

}


//SubClass.ts

class Sub extends Base {
  constructor() {
    super();
  }

  subMethod = () => {
    console.log('Iam subclass method')
  }
}

Is this possible in typescript? the base class complains that subMethod is not defined. This works without the class encapsulation, I have been looking for a way around this as majority of my codebase is in ES6+ syntax

Answer 1

If your Base class implementation relies on a method that is not (and cannot be) defined, then Base is abstract and should be declared as such:

abstract class Base {

  abstract subMethod(): any;

  callSubClassMethod = () => {
    return this.subMethod();
  }

} 

If the method can be defined, so that it performs some default action or returns some default value, then simply define it in Base :

class Base {

  subMethod() {
      // do some default thing, for example, nothing
  }

  callSubClassMethod = () => {
    return this.subMethod();
  }

}