Question

您好我正在尝试获取该声明中属性名称的refence但是当我在fun属性（函数定义）中打印this.name时，我的对象它不起作用，而是当我使用{ {1}}在this.name属性中，它可以正常工作。

为什么？

这是我的代码

this.img

打印未申报....

Answer 1

尝试将scoped存储到变量

中

var submenu = new function(){
    var that = this;
    that.name =  museum[i].name,
    that.title = 'It will merge row',

    that.img ="img/"+that.name + '.png', //it work



     that.fun =  function (data) {
               console.log(that.name); //it doesn't work

     }
 };

修改

发生此错误是因为＆＃34;此＆＃34;是一个范围广泛的关键字，范围的一点变化将影响关键字内容。因此，您将关键字存储在要使用的范围内，而是使用该变量。

Answer 2

因为this个关键字has some funny behaviours，它会根据您调用函数的方式而改变！

var submenu = new function(){
  var self = this;
  //in here, `this` refers to the `submenu` function
  self.name =  museum[i].name;
  self.title = 'It will merge row';
  self.img = 'img/'+ self.name + '.png';

  self.fun =  function (data) {
    console.log(self.name);
    //but in here `this` refers to the `fun` function
  }
};

Answer 3

每个函数调用都有scope and a context与之关联。 Context始终是this关键字的值，它是对“拥有”当前正在执行的代码的对象的引用。尝试将子菜单功能的上下文分配给变量，并在内部函数中使用该变量

var submenu = new function(){
var that = this;
that.name =  museum[i].name,
that.title = 'It will merge row',

that.img ="img/"+that.name + '.png',

that.fun =  function (data) {
           console.log(that.name); 

   }
}

Answer 4

this.name函数中没有this.fun var！请创建一个本地副本，如下所示

   var submenu = new function(){

    this.name =  museum[i].name,
    this.title = 'It will merge row',

    this.img ='img/'+this.name + '.png', //it work

   var clone=this;

     this.fun =  function (data) {
               console.log(clone.name); //it doesn't work

      }
 };

Answer 5

编写您的对象并使用原型设计：

var submenu = function(){
    this.name =  museum[i].name;
    this.title = 'It will merge row';
    this.img ='img/' + this.name + '.png';
};
submenu.prototype.fun =  function (data) {
   console.log(this.name); 
};

在对象中使用此内部函数定义不起作用

5 个答案:

修改