在ES6类中检测静态超级方法

Question

I have one base class and a subclass. I want my class types to have a static method that identifies it by a string. (So I can do a look up in an object for handlers of a type. I can stick the class in it directly, but converts the entire source of the class to a string and uses that as the key, which seems suboptimal.)

I need to:

• detect the presence of a super class
  • call super.id() then append my own id()
• if no super just send my own id()

Here is how I'd imagine the code to be written, but super.id is undefined, so the if always fails. Checking for if (super) {} also fails as a syntax error and super.id() fails as it's "not a function".

class Y {
  static id() {
    if (super.id) {
      return `${super.id()}-${this.name}`;
    }
    else {
      return this.name;
    }
  }
}

class YY extends Y {}

// Outputs "Y YY", but I want "Y Y-YY"
console.log(Y.id(), YY.id()) 

I could define a static id() {} method in YY , but then I'd have to manually do it in all my subclasses which is error prone. Is such a thing possible?

Answer 1

You can use Object.getPrototypeOf instead of super :

 class Y { static id() { if (Object.getPrototypeOf(this).id) { return `${Object.getPrototypeOf(this).id()}-${this.name}`; } else { return this.name; } } } class YY extends Y {} console.log(Y.id(), YY.id()) 

With super it does not work, since that always refers to the prototype of the Y class. But this matches exactly the object you need the prototype of, so with Object.getPrototypeOf(this).id you get a nice bubbling-up through the prototype chain.