ext里如何实现当grid里内容为空返回信息那？

Question

ext里如何实现当grid里内容为空返回信息那？即store没取得数据报个信息到前台
[b]问题补充：[/b]
to yourgame:代码里thiz是啥意思看不明白
[b]问题补充：[/b]
to yourgame:如何扑获store的异常如何返回到前台那

Answer 1

你可以去看下store的api他还有两个事件

exception : ( misc misc )

Fires if an exception occurs in the Proxy during a remote request. This event is relayed through the corresponding Ext.data.DataProxy. See Ext.data.DataProxy.exception for additional details.

loadexception : ( misc misc )

This event is deprecated in favor of the catch-all exception event instead.

This event is relayed through the corresponding Ext.data.DataProxy. See Ext.data.DataProxy.loadexception for additional details.
Listeners will be called with the following arguments:

* misc : misc
  See Ext.data.DataProxy.loadexception for description.

其实我知道你的意思,并不是说捕捉store的什么异常
你的意思肯定是说当你加载数据的时候,后台发生了异常,这个怎么呈现给前台对吧?
我一般是这样的,哪struts2来说,在action里面有一个属性
[code="java"]
public class TestAction extend ActionSupport{
private Map infos = new HashMap();
//setter,getter
public String method(){
if(发生错误){
this.infos.put('msg','错误信息');
}
}
}[/code]

然后配置的时候可以这样来配置
[code="xml"]

true
infos.*

[/code]
然后前台你可以这样来处理
[code="js"]
Ext.Ajax.request({
url: 'ajax_demo/sample.json',
success: function(response, opts) {
var obj = Ext.decode(response.responseText);
alert(obj.infos.msg);//弹出错误
},
failure: function(response, opts) {
console.log('server-side failure with status code ' + response.status);
}
});[/code]