我正在尝试将几个类的静态函数和原型合并到1个主类中,如下面的示例所示。

真的有可能吗?
我的目标是使im能够合并静态变量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

最佳答案

使其简单并使用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实际上是不同的,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带编号的属性。

10-06 12:23