Super in Backbone

When I override the clone() method of a Backbone.Model , is there a way to call this overriden method from my implantation? Something like this:

var MyModel = Backbone.Model.extend({
    clone: function(){
        super.clone();//calling the original clone method
    }
})

---

You'll want to use:

Backbone.Model.prototype.clone.call(this);

This will call the original clone() method from Backbone.Model with the context of this (The current model).

From Backbone docs:

Brief aside on super: JavaScript does not provide a simple way to call super — the function of the same name defined higher on the prototype chain. If you override a core function like set, or save, and you want to invoke the parent object's implementation, you'll have to explicitly call it.

var Note = Backbone.Model.extend({
 set: function(attributes, options) {
 Backbone.Model.prototype.set.apply(this, arguments);
 ...
 }    
});

---

您也可以使用__super__属性，该属性是对父类原型的引用：

var MyModel = Backbone.Model.extend({
  clone: function(){
    MyModel.__super__.clone.call(this);
  }
});

---

Josh Nielsen found an elegant solution for this, which hides a lot of the ugliness.

Just add this snippet to your app to extend Backbone's model:

Backbone.Model.prototype._super = function(funcName){
    return this.constructor.prototype[funcName].apply(this, _.rest(arguments));
}

Then use it like this:

Model = Backbone.model.extend({
    set: function(arg){
        // your code here

        // call the super class function
        this._super('set', arg);
    }
});

链接地址: http://www.djcxy.com/p/18046.html

上一篇: JavaScript依赖管理：npm vs. bower vs. volo

下一篇: 超级骨干