jquery对ajax的error内的XMLHttpRequest返回的exception获取里面的信息

var params = $("form").serialize();
$.ajax({
                    type:"POST",
                    url:"www.xxxxxx.com",
                    data:params,
                    dataType: "json",
                    success:function(data){
                            var backdata = "您提交的姓名为：" + data.name;
for(var i = 0; i<data.posts.length;i++)
{
alert(data.posts[i].id + data.posts[i].title);
break;
}


                            },
                    error:function(XMLHttpRequest, textStatus){                       
                        if(XMLHttpRequest.status==500){
                            var result = eval("("+XMLHttpRequest.responseText+")");
                            alert(result.exception.message);
                          }
                   }
});