[英]Can a instance method in a Javascript subclass call it's parents static method?
这行不通。 JS中的子类是否可以访问其父类的静态方法?
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();
对于静态方法,请使用类名称,而不要使用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();
是的你可以。 您可以使用super
但要获取静态方法,您必须访问它的构造方法属性:
super.constructor.isHuman()
您也可以直接使用类名:
Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();
或通过上链原型链
//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();
演示
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();
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