在jquery通过ajax发送异步请求

1只请求不返回数据

 $.ajax({ type:"post", url:"/ELoan/log/updateFinState.action", data:"finId="+$("#finId").val(), success:function(){ alert("同意融资申请成功"); } });

2，请求并返回数据

$("#Button1").bind("click", function () {                $.ajax({                    type: "post",                    url: "default.aspx",                    data: "name=" + $("#Text1").val(),                    success: function (result) {                        alert(result.msg);                    }                });            });

后台

protected void Page_Load(object sender, EventArgs e)    {        if (Request["name"]!=null)        {            Response.ContentType = "text/json";            Response.Write("{\"msg\":\""+Request["name"]+"\"}");//将数据拼凑为Json            Response.End();        }    }