服务和广播接收器

Question

我看过几个实现BroadCastReceiver的例子，但是如果我需要实现一个必须要响应某些挂起Intent（例如来电）的服务，该如何实现呢？... 实际上，我也在想同样的“问题”，但是在一个Activity中... 显然你有一个类扩展了Service或者Activity，所以它不能再扩展BroadCastReceiver了... 看起来我们无法创建“平台感知”的服务和/或活动？

- TiGer

Answer 1

要注册一个接收特定意图的活动，您需要：

// Flag if receiver is registered 
private boolean mReceiversRegistered = false;

// I think this is the broadcast you need for something like an incoming call
private String INCOMING_CALL_ACTION = "android.intent.action.PHONE_STATE";

// Define a handler and a broadcast receiver
private final Handler mHandler = new Handler();
private final BroadcastReceiver mIntentReceiver = new BroadcastReceiver() {
  @Override
  public void onReceive(Context context, Intent intent) {
    // Handle reciever
    String mAction = intent.getAction();

    if(mAction.equals(INCOMING_CALL_ACTION) {
      // Do your thing   
    }
}

@Override
protected void onResume() {
  super.onResume();

  // Register Sync Recievers
  IntentFilter intentToReceiveFilter = new IntentFilter();
  intentToReceiveFilter.addAction(INCOMING_CALL_ACTION);
  this.registerReceiver(mIntentReceiver, intentToReceiveFilter, null, mHandler);
  mReceiversRegistered = true;
}

@Override
public void onPause() {
  super.onPause();

  // Make sure you unregister your receivers when you pause your activity
  if(mReceiversRegistered) {
    unregisterReceiver(mIntentReceiver);
    mReceiversRegistered = false;
  }
}

然后您还需要在清单文件中添加一个intent-filter：

 <activity android:name=".MyActivity" android:label="@string/name" >
   <intent-filter> 
     <action android:name="android.intent.action.PHONE_STATE" /> 
   </intent-filter>
 </activity>

- m6tt

---

如果您在代码中创建一个新的BroadcastReceiver实例并注册它，同时提供一个意图过滤器对象，那么您就不需要在清单文件中的活动定义中创建一个<intent-filter>。 - hara

Answer 2

实际上，您可以通过向TelephonyManager添加侦听器来对接收到的电话进行反应。

您可以在Service/Activity中定义PhoneStateListener。

private PhoneStateListener mPhoneStateListener = new PhoneStateListener() {

    @Override
    public void onCallStateChanged(int state, String incomingNumber) {
        super.onCallStateChanged(state, incomingNumber);

        switch (state) {
        case TelephonyManager.CALL_STATE_OFFHOOK:
                     break;
        case TelephonyManager.CALL_STATE_RINGING:
                     break;
        case TelephonyManager.CALL_STATE_IDLE:
                     break;
        }
    }

};

然后在onCreate方法中：

mTelephonyManager = (TelephonyManager) getSystemService(TELEPHONY_SERVICE);
mTelephonyManager.listen(mPhoneStateListener, PhoneStateListener.LISTEN_CALL_STATE);

最后，在onDestroy中清除监听器：

mTelephonyManager.listen(mPhoneStateListener, PhoneStateListener.LISTEN_NONE);

在这种情况下，会更加简单。

- Witold Graca

Answer 3

您可以创建一个内部类

class A extends Activity {
 BroadcastReceiver r = new BroadcastReceiver(){
   // code to handle broadcase
 }
}

这个类将会接收事件，你可以将这些事件传递给主处理器，或者直接调用一些外部方法。

- reflog

---

需要使用内部类吗？ - n611x007