数据报套接字未接收答案

【问题标题】：Datagramsocket is not receiving数据报套接字未接收
【发布时间】：2015-01-20 18:55:16
【问题描述】：

大家好，我是java新手，我正在尝试datagramsocket发送和接收数据包。我的客户端正在发送数据包，但服务器没有接收。请帮助我，在此先感谢。
我的客户端代码如下：

outString = "Hello World"  
InetAddress hostAddress = InetAddress.getByName("localhost");
buf = outString.getBytes();
DatagramPacket out = new DatagramPacket(buf, buf.length, hostAddress, 4002);
s.send(out);

我的服务器端代码是：

DatagramSocket sk = new DatagramSocket(PORT);
DatagramPacket dgp = new DatagramPacket(buf, buf.length);
sk = new DatagramSocket(PORT); //PORT is 4002
 while (true) {
    sk.receive(dgp);
    String rcvd = new String(dgp.getData(), 0, dgp.getLength()) + ", from address: "
      + dgp.getAddress() + ", port: " + dgp.getPort();
  System.out.println(rcvd);
}

【问题讨论】：

任何错误信息？如果是，在哪里？

标签： java

【解决方案1】：

这个例子适合我

public static void main(String[] args) throws IOException {
    if (args.length == 0) {
        byte[] buf = new byte[128];
        DatagramPacket dgp = new DatagramPacket(buf, buf.length);
        DatagramSocket sk = new DatagramSocket(PORT); //PORT is 4002
        while (true) {
            sk.receive(dgp);
            String rcvd = new String(dgp.getData(), 0, dgp.getLength()) + ", from address: "
                    + dgp.getAddress() + ", port: " + dgp.getPort();
            System.out.println(rcvd);
        }
    } else {
        String outString = "Hello World";
        InetAddress hostAddress = InetAddress.getByName("localhost");
        byte[] buf = outString.getBytes();
        DatagramPacket out = new DatagramPacket(buf, buf.length, hostAddress, 4002);
        DatagramSocket s = new DatagramSocket();
        s.send(out);
    }
}

【讨论】：