如何根据初始配置设置属性？

Question

我一直在寻找这么久，我找不到具体的答案。比方说，我有一个扩展控制面板从这样的：

Ext.define('MyApp.view.admin.User',{ 
extend: 'Ext.panel.Panel', 
config: { 
    canAdd: false, 
    canDelete: false, 
    canEdit: false 
}, 
constructor: function(config){ 
    this.initConfig(config); 
    this.callParent(arguments); 
}, 
xtype: 'user', 
requires: [ 
    'MyApp.view.admin.UsersGrid', 
    'MyApp.view.admin.UserModel', 
    'MyApp.view.admin.UserController', 
    'MyApp.model.User' 
], 
viewModel: { 
    type: 'user' 
}, 
controller: 'user', 
frame: true, 
layout: { 
    type: 'vbox', 
    align: 'stretch' 
}, 
items: [ 
    { 
     xtype: 'users-grid', 
     flex: 1 
    } 
], 
dockedItems: [ 
    { 
     xtype: 'toolbar', 
     dock: 'top', 
     items: [ 
      { 
       xtype: 'button', 
       text: 'Add', 
       glyph: MyApp.util.Glyphs.getGlyph('add'), 
       hidden: **[config.canAdd]** 
       listeners: { 
        click: 'onAdd' 
       } 
      }, 
      { 
       xtype: 'button', 
       bind: { 
        disabled: '{!usersGrid.selection}' 
       }, 
       text: 'Edit', 
       hidden: **[config.canEdit]** 
       hidden: this.setElementConfiguration, 
       glyph: Mofeg.util.Glyphs.getGlyph('edit'), 
       listeners: { 
        click: 'onEdit' 
       } 
      }, 
      { 
       xtype: 'button', 
       bind: { 
        disabled: '{!usersGrid.selection}' 
       }, 
       text: 'Eliminar', 
       glyph: MyApp.util.Glyphs.getGlyph('destroy'), 
       listeners: { 
        click: 'onDelete' 
       } 
      } 
     ] 
    } 
]}); 

我想申请基于初始配置按钮的隐藏属性，我怎么能做到呢？

Answer 1

你可以写一个initComponent：

initComponent: function() { 
    var me = this, 
     editButton = Ext.ComponentQuery.query('button[itemId=edit]')[0], 
     canEdit = me.getCanEdit(); 

    editButton.setHidden(canEdit); 

    me.callParent(arguments); 
}