ajax 实现局部刷新
点击按钮刷新列表数据 ，求代码
[解决办法]
这个应该使用框架吧...
[解决办法]
求什么代码，显示区域可以套一个div，success返回后，直接根据id，改写id的innerhtml
[解决办法]
比如一下代码，xxxx.ashx中可以处理要现实的内容，并返回给ajax的success,div1为层的id
$.ajax({
type: "POST",
url: "xxxx.ashx",
data:'CarNo='+escape(CarNo)+
'&CarName='+escape(CarName)+
'&CarMark='+escape(CarMark)+
'&Motorman='+escape(Motorman)+
'&Creator='+escape(Creator)+
'&CarType='+escape(CarType)+
'&Factory='+escape(Factory)+
'&FuelType='+escape(FuelType)+
'&EngineNo='+escape(EngineNo)+
'&Status='+escape(Status)+
'&Displacement='+escape(Displacement)+
'&SeatCount='+escape(SeatCount)+
'&Carrying='+escape(Carrying)+
'&BuyMoney='+escape(BuyMoney)+
'&BuyDate='+escape(BuyDate)+
'&VendorName='+escape(VendorName)+
'&VendorAddress='+escape(VendorAddress)+
'&Contact='+escape(Contact)+
'&ContactTel='+escape(ContactTel)+
'&CreateDate='+escape(CreateDate)+
'&Remark='+escape(Remark)+
'&action='+escape(action)+
'&IsNew='+escape(isnew),
dataType:'json',//返回json格式数据
cache:false,
beforeSend:function()
{
AddPop();
},
//complete :function(){hidePopup();},
error: function() {showPopup("../../../Images/Pic/Close.gif","../../../Images/Pic/note.gif","请求发生错误！");},
success:function(data)
{
$("#div1").innerHtml = data;
}
});
