This time I will show you how to solve the status=parsererror error reported during Ajax interaction. What are theprecautionsto solve the status=parsererror error reported during Ajax interaction. The following is a practical case. Let’s take a look. .
Cause: The data returned by the servlet is not in Json format
1. The JS code is:
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!"); } });
2. Note that the above url is /LSHome/LSHome, (the project name is LSHome), so in the web.xml file, configure the Servlet as follows:
LSHomeServlet com.ys.servlet.LSHomeServlet LSHomeServlet /LSHome
3. The code in the Servlet is:
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(); }
4. The result is always an error entering the ajax method, and status=parsererror
xhr = Object {readyState: 4, responseText: "success", status: 200, statusText: "OK"}
5. Solution:
The reason is that the data format returned through the response object is incorrect. The correct method
PrintWriter out = response.getWriter(); String jsonStr = "{\"success\":\"OK\"}"; out.write(jsonStr);
can piece together the return value into JSON data format , and then will it report status=parsererror
I believe you have mastered the method after reading the case in this article. For more exciting information, please pay attention to other related articles on the PHP Chinese website!
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