我试图在android中编写一个应用程序来发布mqtt消息。 我使用AsyncTask发布到我的经纪人,但有时我有一个很大的延迟,我猜是因为我必须每次发布之前连接到经纪人
@Override
protected Void doInBackground(String... params) {
try {
settings = new AppSettings(context);
if (!settings.getBroker().isEmpty()) {
client = new MqttClient(settings.getBroker(), MQTTService.ANDROID_ID, new MemoryPersistence());
if (!settings.getUsername().isEmpty() || settings.getPassword().isEmpty()) {
MqttConnectOptions options = new MqttConnectOptions();
options.setUserName(settings.getUsername());
options.setPassword(settings.getPassword().toCharArray());
client.setCallback(new MQTTPushCallback(context));
client.connect(options);
} else {
client.setCallback(new MQTTPushCallback(context));
client.connect();
}
if (client.isConnected()) {
client.publish(settings.getTopic(), new MqttMessage(params[0].getBytes()));
} else {
Log.d(TAG, "Client is not connected to the mqtt service");
}
} else {
Log.d(TAG, "Broker URL unavailable !");
}
} catch (MqttException e) {
e.printStackTrace();
}
return null;
}
我的问题是,有没有更好的方法来实现这一点,也许在服务中保持连接活着?
最好的问候,
保罗。
答案 0 :(得分:0)
paho网站上有一个Android服务(https://eclipse.org/paho/clients/android/)你应该使用它来保持连接的开放和生存,然后才发布数据
答案 1 :(得分:0)
好的,事实证明它很简单,我所做的就是创建一个子类并扩展BroadcastReceiver
public class PublishBroadcast extends BroadcastReceiver{
@Override
public void onReceive(Context context, Intent intent) {
Log.d(TAG , "onReceive");
if((client != null) && (client.isConnected())){
String Payload = intent.getStringExtra(EXTRA_MESSAGE);
MqttMessage message = new MqttMessage(Payload.getBytes());
try {
client.publish(TOPIC ,message);
} catch (MqttException e) {
e.printStackTrace();
}
}
}
}
比我在服务中的onCreate()我注册了接收器
LocalBroadcastManager.getInstance(getApplicationContext()).registerReceiver(new PublishBroadcast() ,
new IntentFilter(ACTION_PUBLISH));