Question

protected void onCreate(Bundle savedInstanceState) {
    ...
    byte sendData[] = new byte[1024];
    String sentence = "Hey!";
    sendData = sentence.getBytes();
    int ret = 0;
    try {
        DatagramSocket socket = new DatagramSocket();
        InetAddress IPAddress = InetAddress.getByName("192.168.0.11");
        DatagramPacket sendPacket = new 
            DatagramPacket(sendData,4, IPAddress,4202);
        socket.send(sendPacket);
        socket.close();
    }

如果socket.send被注释掉，代码就可以使用。

清单包含： uses-permission android:name="android.permission.INTERNET"

谢谢。

Answer 1

如果你运行socket.send，如果你在onCreate方法或UI线程上运行该代码，它将给出NerworkOnMainThread异常，

尝试创建类似

的线程

  private LinkedBlockingQueue<DatagramPacket> writeQueue = new LinkedBlockingQueue<DatagramPacket>();

private Thread sendThread = new Thread() {
    private boolean shouldStop = false;

    @Override
    public void run() {
        while (!shouldStop) {
            if ((finshed) && (writeQueue.size() == 0)) {
                return;
            }
            try {
                DatagramPacket packet = writeQueue.take();
                if (socket == null) {
                    shouldStop = true;
                } else {
                    socket.send(packet);
                }
            } catch (InterruptedException e) {
                shouldStop = true;
            } catch (IOException e) {
                e.printStackTrace();
                shouldStop = true;
            }
        }
    }
};

UDP，三星S4。应用程序在socket.send（）上停止

1 个答案: