es6类继承

Question

我正在尝试将几个类的静态和原型合并到一个大师类中，如下例所示。

这实际上可行吗？ 我的目标是能够合并静态和原型来执行以下操作：

Alltogether.dosomething（） 或Alltogether.o3.myoption1

// class1
class Option1 {
    constructor(properties) {
        this.o = {};
        this.o.myoption1 = properties.o.myoption1;
        this.o.myoption2 = properties.o.myoption2;
    }
    dosomething1() {
        return "o1";
    }
}

// class2
class Option2 {
    constructor(properties) {
        this.o2 = {};
        this.o2.myoption1 = properties.o2.myoption1;
        this.o2.myoption2 = properties.o2.myoption2;
    }
    dosomething2() {
        return "o2";
    }
}

// class3
class Option3 {
    constructor(properties) {
        this.o3 = {};
        this.o3.myoption1 = properties.o3.myoption1;
        this.o3.myoption2 = properties.o3.myoption2;
    }
    dosomething3() {
        return "o3";
    }
}

// now i want to create a super class with the 3 classes above
// however this doesnt work...
class Alltogether extends (Option1, Option2, Option3) {
    constructor(properties) {
        //........
    }
}

// My goal is that im able to merge the statics and prototypes to do things like
// Alltogether.dosomething()  // o2
// or Alltogether.o3.myoption1

Answer 1

简化并使用composition over inheritance：

class Option {
    constructor(properties) {
        this.myoption1 = properties.myoption1;
        this.myoption2 = properties.myoption2;
    }
    dosomething() {
        return "o";
    }
}

class Alltogether {
    constructor(properties) {
        this.o = new Option(properties.o)
        this.o1 = new Option(properties.o1)
        this.o2 = new Option(properties.o2)
    }
}

如果您的dosomething实际上不同，则可以使用多个Option（子）类。如果没有，您甚至应该考虑使用数组而不是三个编号的属性。