Question

好吧，我是jquery的新手。我创建了一个简单的表单，从jsp页面获取两个数字。 A将它们发送到servlet，该servlet添加并返回总和。我的问题是总和是打印在一个新的空白页面上，但没有显示在index.jsp页面中。我认为jquery没有收到回应。我已阅读并尝试了许多答案，但似乎无法解决问题。另外，如果servlet将任何内容返回给jquery，我该怎么办？ index.jsp页面。

<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Insert title here</title>
<script src="js/jquery-1.11.2.js"></script>
<script>
 var frm = $('#form1');
 frm.submit(function(){

      $.ajax({
           type: frm.attr('method'),
           url:frm.attr('action'),
           data:frm.serialize(),
           dataType:"text",
           success:function(data){
            console.log("success");
            $('#somediv').append('data');
        }


          });
      return false;
       });

</script>
</head>
<body>
<h1 align="center"> ajax with jquery simple example</h1>
<br>
<form name = "form1" action="/ajax_to_servelet/AjaxExample" method="GET" id="form1" accept-charset="UTF-8">
number 1 : <input type="text" name = "n1"/><br>
number 2 : <input type= "text" name  = "n2"/><br>
<input type="submit" value="calculate"/><br>
<div id="somediv"></div></form>
</body>

Servelet的：

/**
 * Servlet implementation class AjaxExample
 */
@WebServlet("/AjaxExample")
public class AjaxExample extends HttpServlet {
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#HttpServlet()
     */
    public AjaxExample() {
        super();
        // TODO Auto-generated constructor stub
    }

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/plain");  // Set content type of the response so that jQuery knows what it can expect.
        response.setCharacterEncoding("UTF-8");
        try{
        int n1 = Integer.parseInt(request.getParameter("n1")) ;
        int n2 = Integer.parseInt(request.getParameter("n2")) ;


        PrintWriter out = response.getWriter();
        out.write(n1+n2+"");
        }catch(NumberFormatException e)
        {
            PrintWriter out = response.getWriter();
            out.write("null exception error"+e);
        }

        // TODO Auto-generated method stub
    }

}

Servlet不在jquery.ajax调用中返回数据

0 个答案: