Solution to the problem that status=parsererror always reports when Servlet interacts with Ajax-JS Tutorial-php.cn

This article mainly introduces the solution to the problem of always reporting status=parsererror when Servlet interacts with Ajax. It is very good and has reference value. Friends in need can refer to it

Reason: 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!"); } });

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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

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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(); }

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4. The result is that it always enters ajax error in the method, and status=parsererror

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

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5. Solution:

The reason is that the data format returned through the response object is incorrect. The correct method is

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

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The above is what I compiled for everyone. I hope it will be helpful to everyone in the future. helpful.

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