如何在JavaScript中创建新对象？

Question

Why is this not working??

    var sheep = function(options){
        this.options = {sizes: 100,
                        eat: 100,
                 colors: 'white', 
                 running: function () {
                     return this.sizes + this.eat;
                 }
          }
    };

    var blacksheep = new sheep({colors:'black'});       

    alert('blackcsheep color is ' + blacksheep.colors);//error undefined
    alert('blackcsheep color is ' + blacksheep.options.colors);// it return white
    alert('blackcsheep running is ' + blacksheep.running());//error

Answer 1

The syntax:

var sheep = {sizes:100, eat:100, colors:'white',running:function(){
        return this.sizes+this.eat;
        }
    };

is an object literal. It defines an instance of an object, but not the class that defines it. Therefore, there is no way to "new-up" another instance of the object.

Take a look at jQuery's extend functionality:

var blacksheep = {
}

$.extend(blacksheep, sheep, { color: black });

This will copy all the properties of sheep into blacksheep , then merge the third parameter into blacksheep , effectively achieving what you want.

Answer 2

要制作另一只基于绵羊的黑羊，在这种情况下你可以做（​​使用jQuery）：

var blacksheep = $.extend(sheep, { color: 'black' });

Answer 3

You can create a sheep object like this.

 function Sheep(sizes,eat,colors){
    this.sizes = sizes;
    this.eat = eat;
    this.colors = colors;
    this.running = function (){
     return this.sizes+this.eat;
    }

    }

Alternatively you can write like this also

 function Sheep(sizes,eat,colors){
    this.sizes = sizes;
    this.eat = eat;
    this.colors = colors;        
    }
 sheep.prototype.running = function(){
 return this.sizes + this.eat;    
}

var sheep1 = new Sheep('100','100','white');

Answer 4

var sheep = function(){
    this.sizes = 100;
    this.eat = 100;
    this.colors = 'white';
    this.running = function(){
        return this.sizers + this.eat;
    }
}

Answer 5

You don't declare objects in JavaScript in the same way as you do in strongly-typed languages. You declare objects by using functions like this:

function sheep() {
    this.color = "white";
    this.size = 200;
    this.speed = 100;
    this.running = function () {
        return "the sheep is running!";
    };
}

var blacksheep = new sheep();

alert('sheep size is ' + blacksheep.size);
alert('sheep running is ' + blacksheep.running());​

---

Your new object does not work because you are creating a new object with a sub-object called options . options contains all of your methods. As such only the second of these three lines that you gave will give you the correct response:

alert('blackcsheep color is ' + blacksheep.colors);
alert('blackcsheep color is ' + blacksheep.options.colors); // Only this one correctly references `options`.
alert('blackcsheep running is ' + blacksheep.running());

Answer 6

in your case sheep is already an object you can not create object of an object. you can directly use that object with property.

But i think you want something like this

var sheep = {sizes:100, eat:100, colors:'white', running:function(){ return this.sizes+this.eat; } }; Object.defineProperty(sheep, 'colors', { value: 'black' , writable: true });

Thanks

Answer 7

Javascript is prototype based not class based. It does not use classes and is also object orientated.

var whitesheep =new sheep("100","100","white","running");
var blacksheep =new sheep("100","100","black","running");