Question

我有一个webservlet，＆＃34; Home.jsp＆＃34;将在打开Web应用程序时加载，因此它将在servlet中调用doGet并将其转发到仪表板jsp页面

@WebServlet("/Home.jsp")
public class HomeController extends HttpServlet {...
....

request.getRequestDispatcher("central_dashboard.jsp").forward(request,response);

因此，在仪表板页面中，我有一个javascript函数，我调用另一个servlet页面＆＃34; Process.do＆＃34;，问题是调用了get方法，并且它没有将页面转发到＆＃34; results.jsp＆＃34;在getRequestDispatcher上。

//Calling the webservlet from the jsp page

function res_call(d, i) {
    $.ajax({  
        type: "GET",  
        url: "Process.do",                             
      });

}

Process.do

request.getRequestDispatcher("results.jsp").forward(request, response);

注意：所有jsp文件都在webcontent文件夹

下

我在这里看过一些帖子，试图在webservlet中返回该函数，但没有用。我不确定问题出在哪里。

发布代码，

   @WebServlet("/Home.jsp")
    public class HomeController extends HttpServlet {
        private static final long serialVersionUID = 1L;

        /**
         * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
         */
        protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
                t

ry {
                doProcessRequest(request, response);
            } catch (SQLException | JSONException | ParseException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
    }
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
            try {
                doProcessRequest(request, response);
            } catch (SQLException | JSONException | ParseException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
    }

    protected void doProcessRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        request.setAttribute("name",val);   
        request.getRequestDispatcher("dashboard.jsp").forward(request,response);
    }
}

    @WebServlet("/Process.do")
    public class ProcessController extends HttpServlet {
        private static final long serialVersionUID = 1L;

        /**
         * @see 

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

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        try {
            doProcessRequest(request, response);
        } catch (SQLException | JSONException | NumberFormatException | ParseException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        try {
            doProcessRequest(request, response);
        } catch (SQLException | JSONException | NumberFormatException | ParseException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
    protected void doProcessRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        request.setAttribute("name",val);   
        request.getRequestDispatcher("results.jsp").forward(request, response);
        return;

    }
}

dashboard.jsp - 从其中包含central.jsp，我将调用js函数，调用它，并执行Process.do - doGet方法，但result.jsp页面没有被转发/打开

<!DOCTYPE html>
<meta charset="utf-8">
    <head>
    <title></title>
    </head>
    <body>
    <div class="centraltest">
     <%@ include file="central.jsp"%>
    </div>
    </body>

Answer 1

1）问题是尝试理解当您从ajax发送http请求时，这意味着您是在单独的线程中而不是在主线程中发送请求（页面本身就是你发送请求的地方）。因此，servlet上的重定向不会反映在客户端。

为了实现这一点，请将要重定向的URL作为对请求的响应发送回来，并在ajax的成功方法上使用java脚本window.location（URL）

内部servlet ：

function res_call(d, i) {
  $.ajax({
      url: "Process",
      type: "GET",
      dataType : 'json',
      cache: false,
      data: { 'url': 'value' }, 
      success: function(data){
          window.location = data.url;
                }
 });
}

在客户端：

window.location.href

注意：当您通过AJAX调用servlet时，无论服务器发送的标头如何，您都可以按照定义停留在同一页面上。

如果您想更改页面，则必须使用javascript，在$ .ajax（..）调用的成功处理函数中执行此操作。

您可以阅读servlet响应发送的位置响应标头，并将response.getOutputStream().print("<script> window.location.href='central_dashboard.jsp'; </script>"); //make sure this is the last line of your program control flow, because from here //the response will be sent to central_dashboard.jsp设置为该值，检查其他options是否有重定向。

2）您可以在处理 servlet的doGet方法中使用以下代码作为替代方法。

request.getRequestDispatcher("central_dashboard.jsp").forward(request,response);

而不是Process servlet中的以下代码：

{"hour": "00", "month": "07", "second": "00", "year": "1970", "timezone": "-00:00", "day": "12", "minute": "00"}
{"hour": "00", "month": "07", "second": "01", "year": "1970", "timezone": "-00:00", "day": "12", "minute": "00"}
{"hour": "00", "month": "07", "second": "02", "year": "1970", "timezone": "-00:00", "day": "12", "minute": "00"}
{"hour": "00", "month": "07", "second": "03", "year": "1970", "timezone": "-00:00", "day": "12", "minute": "00"}
{"hour": "00", "month": "07", "second": "04", "year": "1970", "timezone": "-00:00", "day": "12", "minute": "00"}

无法通过getRequestDispatcher转发到jsp页面

1 个答案: