Javascript子类中的实例方法可以将其称为父方法静态方法吗？

Question

This doesn't work. Is there a way for a child class in JS to have access to its parents static methods?

class Person {
    constructor() {}

    static isHuman() {
        return 'yes I am';
    }
}


class Brian extends Person {
    constructor() {
        super();
    }

    greeting() {
        console.log(super.isHuman())
    }
}

const b = new Brian();
b.greeting();

Answer 1

For static methods, use the class name rather than super.

 class Person { constructor() {} static isHuman() { return 'yes I am'; } } class Brian extends Person { constructor() { super(); } greeting() { console.log(Person.isHuman()) } } const b = new Brian(); b.greeting(); 

Answer 2

Yes you can. You can use super but to get at the static method you have to access the constructor property of it:

super.constructor.isHuman()

You can also use the class name directly:

Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();

Or by going up the prototype chain

//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();

---

Demo

 class Person { constructor() {} static isHuman() { return 'yes I am'; } } class Brian extends Person { constructor() { super(); } greeting() { console.log("From super: ", super.constructor.isHuman()) console.log("From class name (Brian): ", Brian.isHuman()) console.log("From class name (Person): ", Person.isHuman()) console.log("From prototype chain: ", this.constructor.isHuman()) console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman()) } } const b = new Brian(); b.greeting();