能否Android应用程序只播出recevier和服务没有活动? 如果这是可能的我怎么能调用广播接收器? Android系统自动调用broadcsat接收器?
Can android application have only broadcast recevier and service without activity ? If this is possible how can i invoke broadcast receiver ? Android system automatically invokes the broadcsat receiver ?
$ C $的BroadcastReceiver了C
Code of Broadcastreceiver
public class CheckReceiver extends BroadcastReceiver {
public Context con;
@Override
public void onReceive(Context context, Intent intent) {
// TODO Auto-generated method stub
Toast.makeText(context, "Received", Toast.LENGTH_LONG).show();
// add PhoneStateListener
PhoneCallListener phoneListener = new PhoneCallListener();
TelephonyManager telephonyManager = (TelephonyManager) context
.getSystemService(Context.TELEPHONY_SERVICE);
telephonyManager.listen(phoneListener,
PhoneStateListener.LISTEN_CALL_STATE);
con = context;
}
class PhoneCallListener extends PhoneStateListener {
private boolean isPhoneCalling = false;
String LOG_TAG = "LOGGING 123";
@Override
public void onCallStateChanged(int state, String incomingNumber) {
if (TelephonyManager.CALL_STATE_RINGING == state) {
// phone ringing
Log.i(LOG_TAG, "RINGING, number: " + incomingNumber);
}
if (TelephonyManager.CALL_STATE_OFFHOOK == state) {
// active
Log.i(LOG_TAG, "OFFHOOK");
isPhoneCalling = true;
}
if (TelephonyManager.CALL_STATE_IDLE == state) {
// run when class initial and phone call ended, need detect flag
// from CALL_STATE_OFFHOOK
Log.i(LOG_TAG, "IDLE");
if (isPhoneCalling) {
Log.i(LOG_TAG, "restart app");
Intent start = new Intent(con, CheckService.class);
con.startService(start);
isPhoneCalling = false;
}
}
}
}
}
服务code是
Code of Service is
public class CheckService extends Service{
@Override
public IBinder onBind(Intent arg0) {
// TODO Auto-generated method stub
return null;
}
@Override
public void onStart(Intent intent, int startId) {
// TODO Auto-generated method stub
super.onStart(intent, startId);
Intent dialogIntent = new Intent(CheckService.this,SmartDialog.class);
dialogIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
startActivity(dialogIntent);
//Toast.makeText(CheckService.this, "Serive", Toast.LENGTH_LONG).show();
}
}
Android的清单文件
Android Manifest file is
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.mkyong.android"
android:versionCode="1"
android:versionName="1.0" >
<uses-sdk android:minSdkVersion="8" />
<uses-permission android:name="android.permission.CALL_PHONE" />
<uses-permission android:name="android.permission.READ_PHONE_STATE" />
<uses-permission android:name="android.permission.PROCESS_OUTGOING_CALLS" />
<application
android:icon="@drawable/ic_launcher"
android:label="@string/app_name" >
<receiver android:name="CheckReceiver"></receiver>
<service android:name="CheckService"></service>
<activity android:name="SmartDialog" android:theme="@android:style/Theme.Dialog"></activity>
</application>
当我尝试运行此code接收机中,无法启动。任何帮助将是非常有益的。
when i try to run this code receiver is not getting started . any help would be really helpful
与蜂窝开始,BroadcastReceivers安装在停止状态,并且不会触发,直至应用程序已实际运行,即你需要一个活动已至少运行一旦。这是深入解释这个Commonsware博客文章:
Starting with Honeycomb, BroadcastReceivers are installed in a stopped state and will not trigger until an application has actually run, i.e. you will need an activity to have run at least once. This is explained in depth in this Commonsware blog post:
广播回归确认
的Android 3.1