一个面板包含 3 个项目。最后一项有一个事件处理程序附件。在处理程序中,将一个新项目(小部件)添加到父面板。在添加新项目之前,应删除相同 xtype 的旧项目。
这是一个不起作用的示例:
Ext.define('Retroplanner.view.dimension.DimensionMapping', {
extend: 'Ext.form.Panel',
defaultListenerScope: true,
items: [{
xtype: 'component',
html: '<h3> Dimension Mappings</h3>',
},{
xtype: 'bm-separator'
},{
xtype: 'siRemoteCombo',
...
listeners: {
'select': 'onSiRemoteCombo'
}
}
],
onSiRemoteCombo: function(cmb, rec, idx) {
var item;
for(item in this.items){
//here item - is undefined,
//although this.items.length=3, as expected
//alert(item.xtype);
//isXType is not defined for undefined element:
if (item.isXType('propGrid')) {
this.remove(item);
break;
}
}
//the following code works as expected, if the previous is commented
var dimensionMapping = Ext.widget('propGrid');
this.items.add(dimensionMapping);
this.updateLayout();
}
});
我尝试使用索引,但它也不起作用:
Ext.define('Retroplanner.view.dimension.DimensionMapping', {
...
defaultListenerScope: true,
items: [{
xtype: 'component',
...
},{
xtype: 'bm-separator'
},{
xtype: 'siRemoteCombo',
...
listeners: {
'select': 'onSiRemoteCombo'
}
}
],
onSiRemoteCombo: function(cmb, rec, idx) {
//the following code does not remove item in GUI interface.
if (this.items.length == 4)
this.remove(this.items[3], true);
var dimensionMapping = Ext.widget('propGrid');
this.items.add(dimensionMapping);
this.updateLayout();
}
});
我希望能够通过 xtype 删除项目,而不需要任何 id 或其他类型的引用。但如果不可能,那么最好的方法是什么?从其容器中删除 GUI 组件。