Question

我的服务必须通过BradcastReceiver与Activity（MyActivity）通信。这个BroadcastReceiver声明了intosame活动，所以：

    private class MessaggiReceiver extends BroadcastReceiver {
        public static final String ACTION_RELOAD = "it.android.dev.thecode.ACTION_RELOAD";
        @Override
        public void onReceive(Context arg0, Intent intent) {
            if(intent.getAction().equals(ACTION_RELOAD)){
                leggi_act; //Method of MyActivity
            }
        }

    }

但是如何在Manifest中声明MyActivity中包含的brodcast？我在it.android.dev.thecode.MyActivity.MessaggiReceiver中使用了这段代码

<receiver android:name="it.android.dev.thecode.MyActivity.MessaggiReceiver">
            <intent-filter>
                <action android:name="it.android.dev.thecode.ACTION_RELOAD"></action>
            </intent-filter>
</receiver>

但是错了，然后我用it.android.dev.thecode.myactivity $ MessaggiReceiver

使用了这段代码。

<receiver android:name="it.android.dev.thecode.MyActivity$MessaggiReceiver">
                <intent-filter>
                    <action android:name="it.android.dev.thecode.ACTION_RELOAD"></action>
                </intent-filter>
    </receiver>

由于

Answer 1

您无需在清单中注册广播接收器，而是通过活动的onResume中的代码进行注册。不要忘记在onPause中取消注册以避免任何泄漏。

@Override
public void onResume() {
   registerReceiver(mMessageReceiver,
       new IntentFilter("it.android.dev.thecode.ACTION_RELOAD"));
}

@Override
public void onPause() {
   unregisterReceiver(mMessageReceiver);
}

private BroadcastReceiver mMessageReceiver = new BroadcastReceiver() {
  @Override
  public void onReceive(Context context, Intent intent) {
    Log.d("receiver", "Got broadcast: " + intent.getAction());
  }
};

Android：活动和服务沟通

1 个答案: