Question

尝试运行echo服务器时无法读取/写入套接字数据。没有抛出异常，也没有对控制台输入的响应。代码中哪些不正确？

public class Server {

    static ServerSocket server;

    public static void main(String[] args) throws IOException  {
        String hostname = "127.0.0.1";

        try{
            server = new ServerSocket(8888); 
          } catch (IOException e) {
            System.out.println("Could not listen on port 8888");
            System.exit(-1);
          }

        Socket theSocket = null;
        try {
            theSocket = new Socket(hostname, 8888);

            BufferedReader networkIn = new BufferedReader(new InputStreamReader(theSocket.getInputStream()));
            BufferedReader userIn = new BufferedReader(new InputStreamReader(System.in));
            PrintWriter out = new PrintWriter(theSocket.getOutputStream(), true);
            System.out.println("Connected to echo server");

            while (true) {
              String theLine = userIn.readLine();
              if (theLine.equals("."))
                break;
              out.println(theLine);
              out.flush();
              System.out.println("networkIn: "+networkIn.readLine());
            }
            networkIn.close();
            out.close();
            System.out.println("out.close();");
        } catch (IOException e) {
            e.printStackTrace();
        }


    }
}

Answer 1

您创建了两个不相关的套接字，一个用于接受连接，另一个用于连接（主机名，8888）。您需要在服务器套接字上调用accept()以实际连接客户端。请参阅tutorial。

Answer 2

首先添加Socket client = server.accept();。然后，您将看到如何重构其余程序。

无法从套接字的InputStream获取数据

2 个答案: