serverSocket未侦听

Question

serverSocket没有监听本地主机的端口，我尝试了几个端口。即使没有超时线，它也不能工作。请在此代码中提出任何修改建议。

public class server1 extends JApplet implements Serializable{

static JApplet japplet = new JApplet();
private static ServerSocket serverSocket = null;
private static Socket clientSocket = null;
private static final int maxClientsCount = 5;
private static final clientThread[] threads = new clientThread[maxClientsCount];

public void init() {
    tool = Toolkit.getDefaultToolkit();
    setup_applet();
    setup_layout();
    run();
}

public void run() {
    try {
        serverSocket = new ServerSocket(6789);
        serverSocket.setSoTimeout(60000);

    while (true) {
        screen.init_Screen();
        clientSocket = serverSocket.accept();

            int i = 0;
            for (i = 0; i < maxClientsCount; i++) {
                if (threads[i] == null) {
                    (threads[i] = new clientThread(clientSocket, threads))
                            .start();
                    break;
                }
            }
            if (i == maxClientsCount) {
                clientSocket.close();
            }
        }
    } catch (IOException e) {
        System.out.println(e);
    }
}
}

class clientThread extends Thread implements Serializable {

private String clientName = null;
private PrintStream os = null;
private Socket clientSocket = null;
private final clientThread[] threads;
private int maxClientsCount;

public clientThread(Socket clientSocket, clientThread[] threads) {
    this.clientSocket = clientSocket;
    this.threads = threads;
    maxClientsCount = threads.length;
}

public void run() {
    int maxClientsCount = this.maxClientsCount;
    clientThread[] threads = this.threads;

    try {
        os = new PrintStream(clientSocket.getOutputStream());

        while (true) {
            String msg = "server:apl";
            synchronized (this) {
                for (int i = 0; i < maxClientsCount; i++) {
                    if (threads[i] != null && threads[i] == this) {
                        os.println(msg);
                        break;
                    }
                }
            }
            synchronized (this) {
                for (int i = 0; i < maxClientsCount; i++) {
                    if (threads[i] != null && threads[i] != this
                            && threads[i].clientName != null) {
                        BufferedImage image = ImageIO.read(clientSocket.getInputStream());
                        if(image != null) {
                            soms1.screen.paint(image.getGraphics());
                        } else {
                            System.out.println("failed to get");
                        }
                    }
                }
            }
            os.close();
        }
    } catch (IOException e) {
        System.out.println(e);
}
}

我甚至检查了防火墙，如果端口没有被其他进程使用。任何帮助我们都将不胜感激

Answer 1

这并不令人惊讶。您开发的应用程序是一个Java小程序。这意味着它将在web浏览器中运行的通常执行容器。当然，Java浏览器插件必须对用户浏览器中运行的applet设置安全限制。否则，applet将是将恶意代码发送给用户的完美方式。

这种保护机制称为沙箱。这带来了许多限制，在您的上下文中唯一重要的限制是，applet只能打开到它们所来自的主机和端口的网络连接。这意味着它们可以对加载它们的服务器进行HTTP调用。有关详细信息，请参阅此处的文档：http://docs.oracle.com/javase/tutorial/deployment/applet/security.html

但是，您可以像这里解释的那样对小程序进行签名，以摆脱这些限制:对于已签名的小程序，解析现有的清单文件的http://docs.oracle.com/javase/7/docs/technotes/guides/jweb/security/rsa_signing.html。http://docs.oracle.com/javase/7/docs/technotes/guides/jweb/security/manifest.html

长短答案:正确地签署你的applet，你就可以开始工作了。