Model view controller 在MVC视图中显示隐藏项
这是我的视图,当选择“否”时,我想在我的组合框中检索id为“SRInfo”的项目,然后将SRInfo设置为可见:Model view controller 在MVC视图中显示隐藏项,model-view-controller,extjs4,Model View Controller,Extjs4,这是我的视图,当选择“否”时,我想在我的组合框中检索id为“SRInfo”的项目,然后将SRInfo设置为可见: Ext.define('FSSP.view.reqform.newReq', { extend: 'Ext.form.Panel', alias: 'widget.newReqForm', ... items: [ { xtype: 'fieldcontainer', fieldLabel: 'Basic Information', labelSt
Ext.define('FSSP.view.reqform.newReq', {
extend: 'Ext.form.Panel',
alias: 'widget.newReqForm',
...
items: [
{
xtype: 'fieldcontainer',
fieldLabel: 'Basic Information',
labelStyle: 'font-weight:bold;padding:0',
layout: 'anchor',
defaults: {
layout: '100%'
},
fieldDefaults: {
labelAlign: 'top'
},
items: [{
xtype: 'fieldcontainer',
labelStyle: 'font-weight:bold;padding:0',
layout: 'hbox',
defaultType: 'textfield',
fieldDefaults: {
labelAlign: 'top'
},
items:[{
flex: 1,
xtype: 'combo',
fieldLabel: '<font color="red">*</font> SR Open?',
allowBlank: false,
store: new Ext.data.SimpleStore({
data: [['Y', 'Yes'], ['N', 'No']],
id: 0,
fields: ['value', 'text']
}),
listeners: {
scope: this,
'select': function(combo,records){
var i = records[0].get('text');
if( i == 'No'){
})我的问题是,在视图中检索项的正确方法是什么,它有xtype、id和名称。我想我需要调用此方法:child([String selector]):void检索与传递的选择器匹配的此容器的第一个直接子级。传入的选择器必须符合Ext.ComponentQuery选择器。在哪里可以定义选择器?
el.setVisible(true);
}
}
}
...
{
xtype: 'fieldcontainer',
labelStyle: 'font-weight:bold;padding:0',
**id: "SRInfo",**
layout: 'hbox',
defaultType: 'textfield',
hidden: true,
fieldDefaults: {
labelAlign: 'top'
},
items: [
{
flex: 2,
name: 'number',
fieldLabel: 'SR Reason',
allowBlank: false,
margins: '0 0 0 5'
},
{
flex: 2,
name: 'region',
fieldLabel: 'SR Comments',
allowBlank: false,
margins: '0 0 0 5'
}]
}]
},
Ext.define('FSSP.view.reqform.newReq', {
extend: 'Ext.form.Panel',
alias: 'widget.newReqForm',
...
initComponent: function() {
this.hiddenItem = Ext.create(...);
Ext.appy(this,{
items:[
{
...
listener: function(){ this.hiddentItem.show()};
...
},
{
...
},
this,hiddenItem
]