Backbone.js的事件懷疑

Question

我要創建使用backbone.js.Below事件是我js代碼

var Trainee = Backbone.Model.extend(); 

      var TraineeColl = Backbone.Collection.extend({ 
       model: Trainee, 
       url: 'name.json' 
      }); 

      var TraineeView = Backbone.View.extend({ 
       el: "#area", 




       template: _.template($('#areaTemplate').html()), 

       render: function() { 
         this.model.each(function(good){ 
         var areaTemplate = this.template(good.toJSON()); 
         $('body').append(areaTemplate); 
        },this); 

        return this; 
       } 
      }); 

      var good = new TraineeColl(); 
      var traineeView = new TraineeView({model: good}); 
      good.fetch(); 


      good.bind('reset', function() { 
      $('#myButtons').click(function() { 
       traineeView.render(); 
       }); 
      }); 



<div class = "area"></div> 
    <div class="button" id="myButtons"> 
<button class="firstbutton" id="newbutton"> 
    Display 

</button> 
</div> 
    <script id="areaTemplate" type="text/template"> 
       <div class="name"> 
        <%= name %> 
       </div> 
       <div class="eid"> 
        <%= eid %> 
       </div> 
       <div class="subdomain"> 
        <%= subdomain %> 
       </div> 

上點擊顯示按鈕我的O/P是 顯示//這是一個按鈕//

辛杜嘉 E808514 HPS

Shalini E808130 HBS

普里亞 E808515 HSG

從現在的觀點，我必須改變事件綁定到模型中的model..the更改必須到顯示器上顯示點擊按鈕輸出的觀看而觸發。

Answer 1

var TraineeView = Backbone.View.extend({ 
    el: "#area", 

    initialize : function(options){   // you will get the passed model in 
//options.model 
     var trainee = new TraineeColl(); 
     trainee.fetch(); 
     trainee.bind('reset change', this.render,this); //change will trigger render 
// whenever any model in the trainee collection changes or is modified 

    } 

    template: _.template($('#areaTemplate').html()), 

    render: function() { 
      this.model.each(function(trainee){ 
      var areaTemplate = this.template(trainee.toJSON()); 
      $('body').append(areaTemplate); 
      },this); 

     return this; 
    } 
}); 

var traineeView = new TraineeView({model: trainee});     
    }); 

Answer 2

這正是isn't回答你queston但：

如果學員（我已經改名，以研修生）是你應該把它用一個集合：

new TraineeView({collection: trainees}); 

然後在渲染：

this.collection.models.each(function(trainee) 

你propably wan't移動調用視圖外取，在路由器也許：

trainees = new TraineeColl(); 
view = new TraineeView({collection: trainees}); 
trainees.fetch(); 

這樣你的視圖只聽模型。

您也應該移動綁定部分的意見初始化方法

this.collection.bind('reset', function() { 
         this.render(); 
}); 

希望這有助於。