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我试图在 Applet 中获取简单的用户详细信息（姓名、电话号码、性别（选项框））并在 JSP 中显示详细信息。我将所有三个详细信息放在 HashMap 中并将其发送到输出流中。小程序代码如下。

    try 
    {
        userUrl = "http://localhost:8080/AppletTest/display.jsp";

       /* In the web.xml file I have mapped display.jsp to the Servlet */

        testServlet = new URL(userUrl.toString());
        servletConnection = testServlet.openConnection();

        servletConnection.setDoOutput(true);
        servletConnection.setRequestProperty("Content-Type","application/octet-stream");

        ObjectOutputStream oos1 = new ObjectOutputStream(servletConnection.getOutputStream());

        /* DataMap is the HashMap Containing values */

        oos1.writeObject(dataMap);
        oos1.flush();
        oos1.close();

        //  Thread.currentThread().sleep(5000);

    }
    catch(Exception ie)
    {
        ie.printStackTrace();
    }

    // Finally call servlet by going to that page.

    getAppletContext().showDocument(userUrl, "_self");

在 servlet 上，我只需获取 HashMap 并将其转发到 jsp 页面进行显示。

    try
    {
        System.out.println("In Servlet");

        ObjectInputStream inputFromApplet = new ObjectInputStream(request.getInputStream());
        HashMap<String,String> receievedData = (HashMap<String,String>) inputFromApplet.readObject();

        request.setAttribute("dataMap",receievedData);
        request.getRequestDispatcher("display1.jsp").forward(request, response);

        inputFromApplet.close();
    }
    catch (ClassNotFoundException e)
    {
        e.printStackTrace();
    }

正如此处问题中的评论中所问的那样，打印了 Sysout("In Servlet")。但是抛出异常

In Servlet        
SEVERE: Servlet.service() for servlet jsp threw exception java.io.EOFException
at java.io.ObjectInputStream$PeekInputStream.readFully(ObjectInputStream.java:2280)
at java.io.ObjectInputStream$BlockDataInputStream.readShort(ObjectInputStream. java:2749)
at java.io.ObjectInputStream.readStreamHeader(ObjectInputStream.java:779)
at java.io.ObjectInputStream.<init>(ObjectInputStream.java:279)
at org.apache.jsp.display1_jsp._jspService(display1_jsp.java:71)
at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:377)
at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313)
at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
at java.lang.Thread.run(Thread.java:662)

我究竟做错了什么。请帮忙。

4

2 回答 2

0

如果我理解正确，您是：

  1. 向 servlet 发送请求
  2. 从该调用的 e 输入流中读取
  3. 发送一个jsp作为对那个调用的响应
  4. 将用户从 applet 重定向到同一个 servlet
  5. 这会导致第二个请求，其中没有任何序列化的内容
  6. servlet 在第二个请求上失败

不幸的是，这不起作用。您应该使用两个 servlet（或带有可选参数的同一个）来处理这两个请求，一个将从输入流中读取并写入会话，而第二个将从会话中检索并显示在 jsp 中。

于 2011-08-09T00:28:08.147 回答
0

我怀疑小程序抛出了一个异常，而您还没有在 Java 控制台中检测到它。

于 2011-08-09T00:19:49.770 回答