Ajax交互时被报status=parsererror错误如何解决

var jsonStr = {'clusterNum':2,'iterationNum':3,'runTimes':4}; $.ajax({ type: "post", //http://172.22.12.135:9000/Json.json url: "/LSHome/LSHome", dataType : 'json', data : jsonStr, success: function(data,textStatus){ if(textStatus=="success"){ alert("创建任务操作成功"+data); } }, error: function(xhr,status,errMsg){ alert("创建任务操作失败!"); } });

 LSHomeServlet com.ys.servlet.LSHomeServlet   LSHomeServlet /LSHome

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { //聚类数量 String clusterNum = request.getParameter("clusterNum"); //迭代次数 String iterationNum = request.getParameter("iterationNum"); //运行次数 String runTimes = request.getParameter("runTimes"); System.out.println("聚类数量为："+clusterNum+"---迭代次数:"+iterationNum+"---运行次数:"+runTimes); PrintWriter out = response.getWriter(); out.write("success"); out.close(); }

xhr = Object {readyState: 4, responseText: "success", status: 200, statusText: "OK"}

PrintWriter out = response.getWriter(); String jsonStr = "{\"success\":\"OK\"}"; out.write(jsonStr);