Backbone View继承 - 调用parent导致递归

Question

我有一个三个Backbone View类继承:

var preventRecursion = 0;

var parentView = Backbone.View.extend({

  initialize: function(){
    console.log('parentView');
  }
});

var nestedChildView = parentView.extend({

  initialize: function(){

    if (++preventRecursion == 5) {throw "Recursion"};

    console.log('nestedChildView');
    this.constructor.__super__.initialize.apply(this);          
  }
});

var nestedNestedChildView = nestedChildView.extend({

  initialize: function(){
    console.log('nestedNestedChildView');
    this.constructor.__super__.initialize.apply(this);          
  }
});

当我尝试创建nestedNestedChildView时:

var t = new nestedNestedChildView();

我得到无限递归:这是jsfiddle

Answer 1

正如Model.extend上的文档中所述,

简要介绍一下super:JavaScript没有提供一种调用super的简单方法 - 在原型链上定义的同名函数更高.如果覆盖像set或save这样的核心函数,并且想要调用父对象的实现,则必须按以下方式显式调用它:

在类层次结构中,this.constructor始终等于构造函数nestedNestedChildView,这意味着this.constructor.__super__.initialize将是nestedChildView.initialize循环.请参阅http://jsfiddle.net/X5yBb/进行测试.

您可以显式调用类__super __(http://jsfiddle.net/X5yBb/1/)

var nestedChildView = parentView.extend({
  initialize: function(){
    console.log('nestedChildView');
    nestedChildView.__super__.initialize.apply(this);          
  }
});

var nestedNestedChildView = nestedChildView.extend({
  initialize: function(){
    console.log('nestedNestedChildView');
    nestedNestedChildView.__super__.initialize.apply(this);          
  }
});

或者如果您愿意,可以调用原型链上的方法(http://jsfiddle.net/X5yBb/2/):

var nestedChildView = parentView.extend({
  initialize: function(){
    console.log('nestedChildView');
    parentView.prototype.initialize.apply(this);          
  }
});

var nestedNestedChildView = nestedChildView.extend({
  initialize: function(){
    console.log('nestedNestedChildView');
    nestedChildView.prototype.initialize.apply(this);          
  }
});

见访问父类中的骨干和超级骨干关于这一主题进一步信息.