Question

我正在尝试制作一个可以管理更多客户端的简单ECHO服务器。

服务器类：

public class EchoServer {

   protected int          port  ;
   protected ServerSocket socket;
   private   Socket       acceptedSocket;

   public EchoServer(int port) throws IOException {
      this.port = port;
      socket = new ServerSocket(port);
   }

   public void start() throws AcceptingClientException {
      while(!socket.isClosed()) {
         try {
            acceptedSocket = socket.accept();
         }
         catch (IOException e){
            throw new AcceptingClientException();
         }
         ClientHandler ch = new ClientHandler(acceptedSocket);
         ch.run();
      }
   }
}

Runnable客户端处理程序：

public class ClientHandler implements Runnable {

   Socket socket;

   public ClientHandler(Socket socket) {
      this.socket = socket;
   }

   @Override
   public void run() {
      PrintWriter    From_Server = null;
      BufferedReader To_Server   = null;
      String to_server_string    = null;
      try {
         From_Server = new PrintWriter(socket.getOutputStream());
         To_Server   =
            new BufferedReader(
               new InputStreamReader( socket.getInputStream()));
         System.out.println("Stream opened.\n");
         while(true) {
            if(To_Server.ready()){
               System.out.println("Reading input line.\n");
               to_server_string = To_Server.readLine();
               if(to_server_string.equalsIgnoreCase("quit")) {
                  System.out.println("Connection closed on user request.\n");
                  From_Server.print("Bye :)\n");
                  From_Server.close();
                  To_Server.close();
                  socket.close();               
               }
               else {
                  System.out.println(
                     "String '" +
                     to_server_string+"' is not 'quit', echoing.\n");
                  From_Server.print("ECHO: "+to_server_string+"\n");
                  System.out.println("String written on stream, flushing.\n");
                  From_Server.flush();
               }
            }
         }
      }
      catch (IOException e) {
         System.out.println("Stream error (connection closed?).\n");
      }
   }
}

主类

public static void main(String[] args) {
   try {
      EchoServer server= new EchoServer(9999);
      server.start();
   }
   catch (IOException ex) {
      System.out.println("Unable to start server (port is busy?)\n");
      Logger.getLogger(SimpleServer.class.getName()).log(Level.SEVERE, null, ex);
   }
   catch (AcceptingClientException e){
      System.out.println("Unable to accept client\n");
   }
}

多个客户端能够连接到服务器，但ECHO当时只能与一个客户端一起工作（如果我关闭与一个客户端的连接，服务器将自动开始处理另一个客户端），但我可以不明白为什么：当客户端连接到服务器时，用server.accept（）创建的关联socked被传递给一个可运行的客户端处理程序的新实例，该处理程序以handler.run（）启动，服务器应该重新开启在server.accept（）中等待（除非关闭ServerSocket）。我假设问题应该是这个服务器类的方法：

public void start() throws AcceptingClientException {
   while(!socket.isClosed()) {
      try {
         acceptedSocket=socket.accept();
      }
      catch (IOException e){
         throw new AcceptingClientException();
      }
      ClientHandler ch = new ClientHandler(acceptedSocket);
      ch.run();
   }
}

但我无法弄清楚它有什么问题......我错过了什么？

Answer 1

您的代码：

ClientHandler ch = new ClientHandler(acceptedSocket);
ch.run();

不会启动新线程，它会在同一个线程中委托给ClientHandler.run()。

要启动主题，请使用new Thread( ch ).start();，因为ch属于ClientHandler类Runnable。

多线程服务器Java

1 个答案: