Question

我正在尝试使用简单的GUI设置服务器，当客户端建立连接并进行标准操作时，我将在其中进行打印。当我尝试执行它时会发生问题，因为GUI无法启动（即使NetBeans说程序已执行），我也不明白线程之间是否存在某种冲突，或者我是否在其中调用了它。一种机翼方式。这是程序启动的代码：

public class Server extends Application {

@Override
public void start(Stage stage) throws Exception {
    FXMLLoader sLoader = new FXMLLoader(getClass().getResource("server.fxml"));

    BorderPane root = new BorderPane(sLoader.load());

    ServerController sController = sLoader.getController();
    sController.initModel();
    Scene scene = new Scene(root);
    stage.setScene(scene);
    stage.show();
}

这是服务器的控制器，我在这里放在init函数中，即Socket等待客户端的部分：

public class ServerController {

@FXML
private TextArea textarea;

public void initModel() {
    try {
        int i = 1;
        ServerSocket s = new ServerSocket(5000);

        while (true) {
            Socket incoming = s.accept(); // si mette in attesa di richiesta di connessione e la apre
            textarea.setText("In attesa di connessioni...");
            Runnable r = new ThreadedEchoHandler(incoming, i);
            new Thread(r).start();
            i++;
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
}

class ThreadedEchoHandler implements Runnable {

    private Socket incoming;
    private int counter;

    /**
     * Constructs a handler.
     *
     * @param i the incoming socket
     * @param c the counter for the handlers (used in prompts)
     */
    public ThreadedEchoHandler(Socket in, int c) {
        incoming = in;
        counter = c;
    }

    //Con questo gestisco le stringhe
    public void run() {
        try {
            try {
                InputStream inStream = incoming.getInputStream();
                OutputStream outStream = incoming.getOutputStream();

                Scanner in = new Scanner(inStream);
                PrintWriter out = new PrintWriter(outStream, true);

                //out.println("Hello! Enter BYE to exit.");

                // echo client input
                boolean done = false;
                while (!done && in.hasNextLine()) {
                    String line = in.nextLine();
                    //out.println("Echo: " + line);
                    //System.out.println("ECHO: " + line);
                    if (line.trim().equals("BYE")) {
                        done = true;
                    }
                }
            } finally {
                incoming.close();
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

JavaFX和Socket之间的冲突

0 个答案: