java - 为什么我的消息在 Java 套接字服务器中只发送一次？

Question

有一个服务器被认为可以同时为多个客户端提供服务。

所以当客户端连接时，他被添加到客户端数组中。当服务器收到消息时，会将其发送给所有客户端。当一个客户端连接时它工作得很好，但是当我同时有2个客户端时，消息只发送一次，之后就不再工作了。有什么问题吗？

服务器

static DataInputStream inputStream;
static DataOutputStream outputStream;

static ServerSocket serverSocket;
static final int PORT = 3003;

static Socket someClient;

static List<Socket> clients = new ArrayList<>();

public Server()
{
    start();
}

public static void main(String[] args) throws IOException
{
    try{
        serverSocket = new ServerSocket(PORT);

        print("Server started on " + serverSocket.getInetAddress().getHostAddress());

        while (true)
        {
            someClient = serverSocket.accept();

            new Server();
        }

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

@Override
public void run()
{
    try{
        clients.add(someClient);

        print("Connected from " + someClient.getInetAddress().getHostAddress());

        InputStream sin = someClient.getInputStream();
        OutputStream sout = someClient.getOutputStream();

        inputStream = new DataInputStream(sin);
        outputStream = new DataOutputStream(sout);

        String message;

        while (true)
        {
            message = inputStream.readUTF();

            print(message);

            for (int i = 0; i < clients.size(); i++)
            {
                Socket client = clients.get(i);

                OutputStream os = client.getOutputStream();

                DataOutputStream oss = new DataOutputStream(os);

                oss.writeUTF(message);
            }
        }
    } catch (Exception e){
        e.printStackTrace();
    }
}

客户端

socket = new Socket("0.0.0.0", 3003);

        InputStream sin = socket.getInputStream();
        OutputStream sout = socket.getOutputStream();

        inputStream = new DataInputStream(sin);
        outputStream = new DataOutputStream(sout);

        sendButton.addActionListener(new ActionListener() {
            public void actionPerformed(ActionEvent e) {
                if(key != null && key.length() == 16)
                {
                    Date date = new Date();

                    String msg = ">> " + nickname + ": " + messageField.getText()+" | " + date.getHours()+":"+date.getMinutes()+"\n";

                    try {
                        outputStream.writeUTF(Encrypt.AESEncrypt(key, msg));
                    } catch (IOException e1) {
                        e1.printStackTrace();
                    }

                    messageField.setText("");
                }
                else if(key == null)
                    JOptionPane.showMessageDialog(J_Frame, "Your key field is empty");
                else if(key.length() != 16)
                    JOptionPane.showMessageDialog(J_Frame, "Key's length should be 16 symbols");
            }
        });

        while (true)
        {

            String message;

            message = inputStream.readUTF();

            append("\n" + Encrypt.AESDecrypt(key, message));
        }

    } catch (Exception e1) {
        clear();
        append(">> Unable to connect to the server.");
        hideButtons();
    }

Answer 1

每次客户端连接到您的服务器时，它都会替换以前的连接:

while (true)
{
        someClient = serverSocket.accept();
        ...
}

someClient 是静态的:

static Socket someClient;

这意味着它被所有线程共享。此外，对它的访问不会以任何方式同步，这意味着对其值的更改不能保证对其他线程可见。

如Peter Lawrey评论中指出，流也需要是非静态的:

static DataInputStream inputStream;
static DataOutputStream outputStream;

实际上，您始终从“最新”inputStream 中读取的事实可能是您所描述的行为的主要原因。outputStream 似乎未使用，因此最好将其删除。

除此之外，OutputStreams 可能需要刷新才能实际发送数据。