UDP中使用的DatagramPacket

我正在尝试使用Java中的DatagramPacket与服务器和客户端进行通信。我的目标是：当客户说：你好，而服务器说：你好，很高兴见到你。所以我用一个if条件来达到这个目的。但是，当输入为“ Hello”时，它看起来像是if条件（我已经使用过debug）。我想知道为什么会这样吗？看来太奇怪了。谁能帮助我解决这个问题？任何帮助，将不胜感激！谢谢！代码：客户：

public static void main(String[] args) {
        System.out.println("This is Server");
        DatagramSocket server = null;
        byte[] container = new byte[1024];
        try {
            server = new DatagramSocket(9998);
        } catch (SocketException e) {
            e.printStackTrace();
        }
        DatagramPacket dPacket = new DatagramPacket(container, container.length);   
        try {
            while(true) {               
                server.receive(dPacket);
                byte[] datas = dPacket.getData();
                String temp = new String(datas);
                if (temp.equals("Hello")) {
                    System.out.println("Hi~!Nice to meet you!");
                }else if (temp.equals("How are you?")) {
                    System.out.println("I am good!");
                }else {
                    System.out.println("Have a good day");
                }
                if (temp.equals("exit")) {
                    break;
                }


            }
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        server.close();
    }

客户：

public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println("This is Client");
        try {
            DatagramSocket client = new DatagramSocket(8888);
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            String input = "";
            while(!input.equals("exit")) {
                input = br.readLine();
                byte[] msg = input.getBytes();
                DatagramPacket dPacket = new DatagramPacket(msg, msg.length,new InetSocketAddress("localhost",9998));
                client.send(dPacket);
            }


            client.close();
        } catch (Exception e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

问题来源：Stack Overflow