Emberjs, each block helper advanced

Question

This is my flat model:

 [
    { id: 1, title: '', body: '' },
    { id: 2, title: '', body: '' },
    { id: 3, title: '', body: '' },
    { id: 4, title: '', body: '' },
    // ...
 ]

This is my expected template:

  <div class="row">
    <div class="col-md-4 col-md-offset-2">
      <h6>{{! item1 title }}</h6>
      <p>{{! item1 body}}</p>
    </div>
    <div class="col-md-4 col-md-offset-1">
      <h6>{{! item2 title}}</h6>
      <p>{{! item2 body}}</p>
    </div>
  </div>
  <div class="row">
    <div class="col-md-4 col-md-offset-2">
      <h6>{{! item3 title }}</h6>
      <p>{{! item3 body}}</p>
    </div>
    <div class="col-md-4 col-md-offset-1">
      <h6>{{! item4 title}}</h6>
      <p>{{! item4 body}}</p>
    </div>
  </div>
  <!-- ... -->

What is the ember approach to do this?

A simple each helper won't help because template combines two elements from the model.

Should I change my model, customize an array controller, build a component, write a template helper, or is there a built in solution for this simple problem?

Answer 1

You should change your data structure. I'd still use an ArrayController, but at the same time I'd add a computed property on it that grouped everything up (I'm guessing your pattern is every other).

App.FooController = Em.ArrayController.extend({
  groupedItems: function(){
    var groups = [],
        group,
        model = this.get('model'),
        len = this.get('length');
    for(var i = 0;i<len;i++){
      if(i%2==0){
        group = [];
        groups.push(group);
      }
      group.push(model[i]);
    }
    return groups;
  }.property('model.[]')
});

{{#each item in groupedItems}}
  <div class="row">
    <div class="col-md-4 col-md-offset-2">
      <h6>{{ item.firstObject.title }}</h6>
      <p>{{ item.firstObject.body}}</p>
    </div>
    <div class="col-md-4 col-md-offset-1">
      <h6>{{item.lastObject.title}}</h6>
      <p>{{item.lastObject.body}}</p>
    </div>
  </div>
{{/each}} 

Honestly, there are a slew of ways of grouping them, the point I'm trying to make is grouping would be the simplest.