如何基於現有類創建新的JavaScript類並將其擴展

Question

我正在學習JavaScript，對於如何基於現有類創建新的JavaScript類並對其進行擴展有點困惑。

我有一個Person類，我創建了2個實例（John和Suzie）。現在，我想創建另一個基於Person的類（Employee），只有Employee Number等更多屬性（並且可以返回John以便從Employee類繼承）。

這是我有：

var Person = function (name, age){ 
    this.name = name || "UNKNOWN"; 
    this.age= age || "UNKNOWN"; 
} 
Person.prototype.sayHello = function() { 
    console.log("Hello, my name is " + this.name + " and I am " + this.age + " years old"); 
}; 
var john = new Person("John Smith",72); 
var Suzie = new Person("Suzie Brown", 25); 

Answer 1

將您的原型，以你的父母的實例，請確保調用在孩子的構造函數的父類的構造：

var Person = function (name, age){ 
    this.name = name || "UNKNOWN"; 
    this.age= age || "UNKNOWN"; 
} 

Person.prototype.sayHello = function() { 
    console.log("Hello, my name is " + this.name + " and I am " + this.age + " years old"); 
}; 

var Employee = function(name, age, id) { 
    Person.call(this, name, age); 

    this.id = id || 'UNKNOWN'; 
}; 

Employee.prototype = new Person(); 

Employee.prototype.getHired = function() { 
    console.log('ZOMG I GOT HIRED! My ID # is:', this.id); 
}; 

使用它的一些示例：

var bob = new Person('Bobby', 25); 
console.log(bob.name); //Bobby 
console.log(bob.age); //25 
console.log(bob.id); //undefined 
bob.sayHello(); //Hello, my name is Bobby and I am 25 years old 

var suse = new Employee('Susan', 32, 1337); 
console.log(suse.name); //Susan 
console.log(suse.age); //32 
console.log(suse.id); //1337 
suse.sayHello(); //Hello, my name is Susan and I am 32 years old 
suse.getHired(); //ZOMG I GOT HIRED! My ID # is: 1337 

Answer 2

//create Employee 
var Employee = function(name, age, ...extra parameters){ 
    this.name = name || "UNKNOWN"; 
    this.age= age || "UNKNOWN"; 
    //employee related stuff... 
}; 
//extend Person 
Employee.prototype = new Person(); 

Answer 3

var Employee = function (name, age, employeeNumber) { 
    ... 
}; 

Employee.prototype = new Person(); // copies Person's name and age to Employee 
Employee.prototype.employeeNumber = 0; // or whatever defaults you want 

var Bob = new Employee(...);