java - 具有循环消息传递的简单客户端服务器

Question

我正在尝试实现一个简单的客户端服务器程序，它将不断交换消息，直到客户端决定停止。我发现了很多关于这个主题的教程，但是我在正确实现循环方面遇到了困难。服务器处理第一个请求，但不处理其他请求。

这可能是一些愚蠢的错误，所以请原谅我问这样的基本问题 - 我是套接字新手。我很乐意提供任何帮助。我提供了所有代码(基于我找到的一些示例):

客户:

public class Client {
    public static void main(String argv[]) throws Exception {

        talkWithServer();
    }

    private static void talkWithServer() throws UnknownHostException, IOException {
        String sentence;
        String serverResponse;
        BufferedReader brClient = new BufferedReader(new InputStreamReader(System.in));

        Socket clientSocket = new Socket("localhost", 9000);
        DataOutputStream out = new DataOutputStream(clientSocket.getOutputStream());
        BufferedReader brServer = new BufferedReader(new InputStreamReader(clientSocket.getInputStream()));

        while(true) {
            sentence = brClient.readLine();
            out.writeBytes(sentence + '\n');
            serverResponse = brServer.readLine();
            System.out.println(serverResponse);

            if (serverResponse.contains("<BYE>")) {
                break;
            }
        }

        clientSocket.close();
    }
}

服务器:

public class Server {
    public static void main(String args[]) throws Exception {
        String clientSentence;
        ServerSocket welcomeSocket = new ServerSocket(9000);

        Protocol protocol = new Protocol();

        while (true) {
            Socket connectionSocket = welcomeSocket.accept();
            BufferedReader inFromClient = new BufferedReader(
                    new InputStreamReader(connectionSocket.getInputStream()));
            DataOutputStream outToClient = new DataOutputStream(
                    connectionSocket.getOutputStream());

            clientSentence = inFromClient.readLine();

            String response = protocol.processInput(clientSentence);

            outToClient.writeBytes(response + '\n');
        }
    }
}

协议(protocol):

public class Protocol {

    public String processInput(String theInput) {
        String theOutput = "> " + theInput;


        return theOutput;
    }
}

为了更容易调试，我简化了示例。感谢您的任何提示!

Answer 1

我的猜测是“Socket connectionSocket =welcomeSocket.accept();”行如果我没记错的话，这将尝试每次接受新的客户端，并且由于您只连接一个，它将在第二次迭代中永远等待该线路。我建议您将该行粘贴到 while 循环之前。