Extjs将动态创建的面板对象添加到现有面板的位置

Question

In my code, I create a panel object like below:

var grid = Ext.create('Ext.grid.Panel', {});

Then I am trying to define a widget which should contain the above panel as one of it's items. I got the definition of my widget working fine. But how can I include the above grid object so that it is rendered as part of the widget:

Ext.define('Ext.simplegridwidget.GridTest', {
    extend: 'Ext.panel.Panel',
    alias: 'widget.gridtest',
    title: 'User Management'
});

Answer 1

An even better approach so that everything related to my widget stays within the class definition...

  Ext.define('Ext.simplegridwidget.GridTest', {
        extend: 'Ext.panel.Panel',
        alias: 'widget.gridtest',
        title: 'User Management',
        initComponent: function() {

            var grid = Ext.create('Ext.grid.Panel', {});

            Ext.apply(this, {
               items: [ grid, details]
           });

           return this.callParent(arguments);
        }
    });

Answer 2

Ok.. figured it out. Not too difficult after all...

   var grid = Ext.create('Ext.grid.Panel', {});

   Ext.define('Ext.simplegridwidget.GridTest', {
        extend: 'Ext.panel.Panel',
        alias: 'widget.gridtest',
        title: 'User Management',
        items: [grid]
    });