Java 在android中的数据报套接字上发送和接收UDP数据包
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sending and receiving UDP packet on datagram socket in android
提问by Ameer Humza
I have two classes,one sender class and the other is the receiver class.Both of the sending and receiving apps stops after few seconds and close down. My sender class is :
我有两个类，一个是发送者类，另一个是接收者类。发送和接收应用程序都在几秒钟后停止并关闭。我的发件人类是：
public class MainActivity extends Activity {
InetAddress receiverAddress;
DatagramSocket datagramSocket;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
try {
datagramSocket = new DatagramSocket(4444);
} catch (SocketException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
byte[] buffer = "0123456789".getBytes();
byte[] address="192.168.1.101".getBytes();
try {
receiverAddress = InetAddress.getByAddress(address);
} catch (UnknownHostException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
DatagramPacket packet = new DatagramPacket(
buffer, buffer.length, receiverAddress, 4444);
try {
datagramSocket.send(packet);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
My receiving or listening class is:
我的接收或听力课程是：
public class MainActivity extends Activity {
DatagramSocket datagramSocket;
DatagramPacket packet;
TextView tv1;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
tv1=(TextView)findViewById(R.id.textView1);
try {
datagramSocket = new DatagramSocket(80);
} catch (SocketException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
byte[] buffer = new byte[10];
packet = new DatagramPacket(buffer, buffer.length);
try {
datagramSocket.receive(packet);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
byte[] buff = packet.getData();
tv1.setText(buff.toString());
}
Thanks in advance for the help.
在此先感谢您的帮助。
回答by sleinen
The port numbers in the "new DatagramSocket(...)" calls look weird. The client should create an "unbound" socket - simply use "new DatagramSocket();". The sender should bind to the port that the client sends to, i.e. "new DatagramSocket(4444);".
“new DatagramSocket(...)”调用中的端口号看起来很奇怪。客户端应该创建一个“未绑定”的套接字——只需使用“new DatagramSocket();”。发送方应该绑定到客户端发送到的端口，即“new DatagramSocket(4444);”。
回答by sleinen
Source and destination port number should be same. Give same numbers in "DatagramSocket(xxx)". xxx must be same in both programs.
源端口号和目标端口号应该相同。在“DatagramSocket(xxx)”中给出相同的数字。两个程序中的 xxx 必须相同。
回答by Sven
In Android you're not allowed to execute Network operations on the UIThread (Main-Thread)
在 Android 中，您不允许在 UIThread（主线程）上执行网络操作
To Fix this: Copy your network-code to a new Thread and let it run.
要解决此问题：将您的网络代码复制到一个新线程并让它运行。
