如何使用UDP套接字在Android的？

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我想使用UDP套接字在Android的，在这里我把我的字符串从Android模拟器，并接收通过电脑我的Java程序，但我的Java程序不会收到任何东西，虽然当我用Java程序的客户端和服务器（我做了两个不同的Java程序），它的工作。

这是我的Andr​​oid主要活动：

公共类首先扩展活动{ 按钮B; @覆盖 公共无效的onCreate（包savedInstanceState）{     super.onCreate（savedInstanceState）;     的setContentView（R.layout.main）;     B =（按钮）findViewById（R.id.button1）;     b.setOnClickListener（新View.OnClickListener（）{         公共无效的onClick（视图v）{             // TODO自动生成方法存根             线程t =新主题（新二（））;             t.start（）;         }     }）;

下面是我的第二类机器人：

公共类二实现Runnable { 第二（） {     跑（）; } 公共无效的run（）{     // TODO自动生成方法存根     尝试 {         字符串messageStr =你好Android的！;         INT SERVER_PORT = 9876;         DatagramSocket的S =新的DatagramSocket（）;         InetAddress类本地= InetAddress.getByName（127.0.0.1）;         INT msg_length = messageStr.length（）;         byte []的消息= messageStr.getBytes（）;         DatagramPacket类P =新的DatagramPacket（消息，msg_length，地方，                 SERVER_PORT）;         s.send（对）;     }赶上（例外五）{     } } }

这是在PC上我的Java code：

公共静态无效的主要（字符串的args []）抛出异常     {             DatagramSocket类的ServerSocket =新的DatagramSocket（9876）;             byte []的receiveData =新的字节[1024];             byte []的送出数据=新的字节[1024];             而（真）             {                     DatagramPacket类receivePacket =新的DatagramPacket（receiveData，receiveData.length）;                     serverSocket.receive（receivePacket）;                     串句=新的String（receivePacket.getData（），0，receivePacket.getLength（））;                     InetAddress类ip地址= receivePacket.getAddress（）;                     INT端口= receivePacket.getPort（）;                     的System.out.println（收到消息+句子++ ip地址++端口）;             }      }

解决方案

在你的Andr​​oid仿真（和Android设备），127.0.0.1意味着Android的模拟机，而不是PC主机。您可以在10.0.2.2

I am trying to use UDP sockets in a android, here I send my string from android emulator and receive that by my Java program on PC, but my Java program does not receive anything, although when I used Java program as both client and server (I made two different Java programs) it worked.

This is my android main activity :

public class First extends Activity { Button b; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); b = (Button) findViewById(R.id.button1); b.setOnClickListener(new View.OnClickListener() { public void onClick(View v) { // TODO Auto-generated method stub Thread t = new Thread(new Second()); t.start(); } });

Here is my second class in android :

public class Second implements Runnable { Second() { run(); } public void run() { // TODO Auto-generated method stub try { String messageStr = "Hello Android!"; int server_port = 9876; DatagramSocket s = new DatagramSocket(); InetAddress local = InetAddress.getByName("127.0.0.1"); int msg_length = messageStr.length(); byte[] message = messageStr.getBytes(); DatagramPacket p = new DatagramPacket(message, msg_length, local, server_port); s.send(p); } catch (Exception e) { } } }

This is my Java code on PC:

public static void main(String args[]) throws Exception { DatagramSocket serverSocket = new DatagramSocket(9876); byte[] receiveData = new byte[1024]; byte[] sendData = new byte[1024]; while(true) { DatagramPacket receivePacket = new DatagramPacket(receiveData,receiveData.length); serverSocket.receive(receivePacket); String sentence = new String(receivePacket.getData(),0,receivePacket.getLength()); InetAddress IPAddress = receivePacket.getAddress(); int port = receivePacket.getPort(); System.out.println("MESSAGE RECEIVED "+sentence+" "+IPAddress+" "+port); } }

解决方案

On your Android emulation (and Android device), 127.0.0.1 means the Android emulation machine, not the host PC. You can access your host at 10.0.2.2