尝试将小程序与 servlet 通信的协议错误

Question

当我将我的名字从 applet 发送到 servlet 时，我遇到了“java.net.UnknownServiceException：协议不支持输出”问题。 我试过Sending object to Servlet throws an error I cant resolve，但没有解决我的问题。 我的小程序代码是：

import java.io.*;

import javax.servlet.ServletException;
import javax.servlet.http.*;

public class AppletToServlet extends HttpServlet {
// Getting a String object from the applet and send it back.
    public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        try {
            response.setContentType("application/x-java-serialized-object");
            InputStream inputStream = request.getInputStream();
            ObjectInputStream inputFromApplet = new ObjectInputStream(inputStream);
            String string = (String) inputFromApplet.readObject();
// getting string value and passing to applet
OutputStream outputStream = response.getOutputStream();
ObjectOutputStream objectOutputStream = new ObjectOutputStream(outputStream);
objectOutputStream.writeObject(string);
objectOutputStream.flush();
objectOutputStream.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

servlet 代码是：

import java.applet.Applet;
import java.awt.*;
import java.awt.event.*;
import java.io.*;
import java.net.*;

public class AppletToServletExample extends Applet {
  private TextField inputField = new TextField(10);
  private TextField resultField = new TextField(10);

  public void init() {
    // add input label, field and send button
    add(new Label("Input Your Name", Label.RIGHT));
    add(inputField);
    Button sendButton = new Button("Send");
    add(sendButton);
    sendButton.addActionListener(new ActionListener() {
      public void actionPerformed(ActionEvent e) {
        SendData();
      }
    });
    // add output label as a non editable field
    add(new Label("Output:", Label.RIGHT));
    add(resultField);
    resultField.setEditable(false);
  }

  // Get a connection to the servlet.
  private URLConnection getServletConnection() throws MalformedURLException,
      IOException {
    URL urlServlet = new URL(getCodeBase(), "applettoservlet");
    URLConnection con = urlServlet.openConnection();
    con.setDoInput(true);
    con.setDoOutput(true);
    con.setUseCaches(false);
    con.setRequestProperty("Content-Type",
        "application/x-java-serialized-object");
    return con;
  }

  // Send the inputField data to the servlet and show the result in the
  // outputField.
  private void SendData() {
    try {
      String input = inputField.getText();
      // send data to the servlet
      URLConnection con = getServletConnection();
      OutputStream outputStream = con.getOutputStream();
      ObjectOutputStream oos = new ObjectOutputStream(outputStream);
      oos.writeObject(input);
      oos.flush();
      oos.close();
      // receive result from servlet
      InputStream inputStream = con.getInputStream();
      ObjectInputStream inputFromServlet = new ObjectInputStream(
          inputStream);
      String result = (String) inputFromServlet.readObject();
      inputFromServlet.close();
      inputStream.close();
      // show result
      resultField.setText(result);
    } catch (Exception ex) {
      ex.printStackTrace();
    }
  }
}

我几乎到处搜索，但无法解决。

Answer 1

您在哪里运行小程序？您是否真的将它嵌入到页面中并通过某种网络服务器请求它？

代码

 URL urlServlet = new URL(getCodeBase(), "applettoservlet");

只有在您从 Web/应用程序服务器提供的页面运行小程序时才会按预期工作。 尝试添加

System.out.println(urlServlet);

检查正在使用的网址。