如何检查接收器是否在Android中注册？

Question

我需要检查我的注册接收器是否仍在注册中，如果没有，我该如何检查任何方法？

Answer 1

如果您考虑的话，我不确定API直接提供API 这个线程:

我想知道同样的事情。
就我而言，我有一个 BroadcastReceiver 呼叫的实现 Context#unregisterReceiver(BroadcastReceiver) 在处理收到的意图后，将自己作为论点传递。
接收器的可能性很小 onReceive(Context, Intent) 方法被多次调用，因为它已注册为多个 IntentFilters, ，创造潜力 IllegalArgumentException 被扔掉 Context#unregisterReceiver(BroadcastReceiver).

就我而言，我可以存储一个私人同步成员以进行检查之前 Context#unregisterReceiver(BroadcastReceiver), ，但是如果API提供了检查方法，则会更加干净。

Answer 2

没有API函数可以检查接收器是否已注册。解决方法是将您的代码放在 try catch block as done below.

try {

 //Register or UnRegister your broadcast receiver here

} catch(IllegalArgumentException e) {

    e.printStackTrace();
}

Answer 3

最简单的解决方案

在接收器中：

public class MyReceiver extends BroadcastReceiver {   
    public boolean isRegistered;

    /**
    * register receiver
    * @param context - Context
    * @param filter - Intent Filter
    * @return see Context.registerReceiver(BroadcastReceiver,IntentFilter)
    */
    public Intent register(Context context, IntentFilter filter) {
        try {
              // ceph3us note:
              // here I propose to create 
              // a isRegistered(Contex) method 
              // as you can register receiver on different context  
              // so you need to match against the same one :) 
              // example  by storing a list of weak references  
              // see LoadedApk.class - receiver dispatcher 
              // its and ArrayMap there for example 
              return !isRegistered 
                     ? context.registerReceiver(this, filter) 
                     : null;
            } finally {
               isRegistered = true;
            }
    }

    /**
     * unregister received
     * @param context - context
     * @return true if was registered else false
     */
     public boolean unregister(Context context) {
         // additional work match on context before unregister
         // eg store weak ref in register then compare in unregister 
         // if match same instance
         return isRegistered 
                    && unregisterInternal(context);
     }

     private boolean unregisterInternal(Context context) {
         context.unregisterReceiver(this); 
         isRegistered = false;
         return true;
     }

    // rest implementation  here 
    // or make this an abstract class as template :)
    ...
}

在代码中：

MyReceiver myReceiver = new MyReceiver();
myReceiver.register(Context, IntentFilter); // register 
myReceiver.unregister(Context); // unregister 

广告1

- 回复：

这确实不是那么优雅，因为您必须记住在注册后设置iSnegistered标志。 - 隐形拉比

- “更多的Ellegant方法”在接收器中添加的方法以注册并设置标志

如果您重新启动设备或您的应用被OS杀死，这将无法正常工作。 - Amin 6小时前

@AMIN-请参阅《代码的生命周期》（未经清单条目注册的系统）注册接收器:)

Answer 4

我正在使用这个解决方案

public class ReceiverManager {

    private static List<BroadcastReceiver> receivers = new ArrayList<BroadcastReceiver>();  
    private static ReceiverManager ref;
    private Context context;

    private ReceiverManager(Context context){
        this.context = context;
    }

    public static synchronized ReceiverManager init(Context context) {      
        if (ref == null) ref = new ReceiverManager(context);
        return ref;
    }

    public Intent registerReceiver(BroadcastReceiver receiver, IntentFilter intentFilter){
        receivers.add(receiver);
        Intent intent = context.registerReceiver(receiver, intentFilter);
        Log.i(getClass().getSimpleName(), "registered receiver: "+receiver+"  with filter: "+intentFilter);
        Log.i(getClass().getSimpleName(), "receiver Intent: "+intent);
        return intent;
    }

    public boolean isReceiverRegistered(BroadcastReceiver receiver){
        boolean registered = receivers.contains(receiver);
        Log.i(getClass().getSimpleName(), "is receiver "+receiver+" registered? "+registered);
        return registered;
    }

    public void unregisterReceiver(BroadcastReceiver receiver){
        if (isReceiverRegistered(receiver)){
            receivers.remove(receiver);
            context.unregisterReceiver(receiver);
            Log.i(getClass().getSimpleName(), "unregistered receiver: "+receiver);
        }
    }
}

Answer 5

您有几个选择

1. 您可以在课堂或活动中贴上标志。将布尔变量放入您的课程中，并查看此标志，以了解您是否已注册接收器。

2. 创建一个扩展接收器的类，您可以使用：

   1. Singleton模式仅在您的项目中具有一个实例。

   2. 实施方法以了解接收器是否为寄存器。

Answer 6

您必须使用try/catch：

try {
    if (receiver!=null) {
        Activity.this.unregisterReceiver(receiver);
    }
} catch (IllegalArgumentException e) {
    e.printStackTrace();
}

Answer 7

你可以轻松做到...

1）创建一个布尔变量...

private boolean bolBroacastRegistred;

2）注册广播接收器时，将其设置为true

...
bolBroacastRegistred = true;
this.registerReceiver(mReceiver, new IntentFilter(BluetoothDevice.ACTION_FOUND));
....

3）在Onpause（）中做...

if (bolBroacastRegistred) {
    this.unregisterReceiver(mReceiver);
    bolBroacastRegistred = false
}

只是它，现在，您将不会在Onpause（）上收到更多异常错误消息。

提示1：始终在ondestroy（）tip2中使用onpause（）中的unregisterReceiver（）：不要忘记将bolbroadcastregistred变量设置为false，当运行unnegisterReceive（）

成功！

Answer 8

如果您将其放在Ondestroy或Onstop方法上。我认为，当活动再次创建后，没有创建MessageReciver。

@Override 
public void onDestroy (){
    super.onDestroy();
LocalBroadcastManager.getInstance(context).unregisterReceiver(mMessageReceiver);

}

Answer 9

我使用意图让广播接收器知道主活动线程的处理程序实例，并使用消息将消息传递给主要活动

我已经使用这种机制检查广播接收器是否已经注册。有时，当您动态注册广播接收器并且不想将其进行两次，或者如果广播接收器正在运行，则需要它。

主要活动：

public class Example extends Activity {

private BroadCastReceiver_example br_exemple;

final Messenger mMessenger = new Messenger(new IncomingHandler());

private boolean running = false;

static class IncomingHandler extends Handler {
    @Override
    public void handleMessage(Message msg) {
        running = false;    
        switch (msg.what) {
        case BroadCastReceiver_example.ALIVE:
    running = true;
            ....
            break;
        default:

            super.handleMessage(msg);
        }

    }
    }

@Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

    IntentFilter filter = new IntentFilter();
        filter.addAction("pl.example.CHECK_RECEIVER");

        br_exemple = new BroadCastReceiver_example();
        getApplicationContext().registerReceiver(br_exemple , filter); //register the Receiver
    }

// call it whenever you want to check if Broadcast Receiver is running.

private void check_broadcastRunning() {    
        /**
        * checkBroadcastHandler - the handler will start runnable which will check if Broadcast Receiver is running
        */
        Handler checkBroadcastHandler = null;

        /**
        * checkBroadcastRunnable - the runnable which will check if Broadcast Receiver is running
        */
        Runnable checkBroadcastRunnable = null;

        Intent checkBroadCastState = new Intent();
        checkBroadCastState .setAction("pl.example.CHECK_RECEIVER");
        checkBroadCastState .putExtra("mainView", mMessenger);
        this.sendBroadcast(checkBroadCastState );
        Log.d(TAG,"check if broadcast is running");

        checkBroadcastHandler = new Handler();
        checkBroadcastRunnable = new Runnable(){    

            public void run(){
                if (running == true) {
                    Log.d(TAG,"broadcast is running");
                }
                else {
                    Log.d(TAG,"broadcast is not running");
                }
            }
        };
        checkBroadcastHandler.postDelayed(checkBroadcastRunnable,100);
        return;
    }

.............
}

广播接收器：

public class BroadCastReceiver_example extends BroadcastReceiver {


public static final int ALIVE = 1;
@Override
public void onReceive(Context context, Intent intent) {
    // TODO Auto-generated method stub
    Bundle extras = intent.getExtras();
    String action = intent.getAction();
    if (action.equals("pl.example.CHECK_RECEIVER")) {
        Log.d(TAG, "Received broadcast live checker");
        Messenger mainAppMessanger = (Messenger) extras.get("mainView");
        try {
            mainAppMessanger.send(Message.obtain(null, ALIVE));
        } catch (RemoteException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
    .........

}

}

Answer 10

我个人使用了调用UnregisterReceiver的方法，并吞下例外情况。我同意这是丑陋的，但目前提供的最佳方法。

我提出了一个功能请求，以获取布尔方法，以检查是否已将接收器添加到Android API中。如果您想添加它，请在此处支持它：https://code.google.com/p/android/issues/detail?id=73718

Answer 11

我遇到了您的问题，我在应用程序中遇到了同样的问题。我在应用程序中多次调用registerReceiver（）。

解决此问题的一个简单解决方案是在您的自定义应用程序类中调用registerReceiver（）。这将确保您的广播接收器将在整个应用程序生命周期中仅被称为一个。

public class YourApplication extends Application
{
    @Override
    public void onCreate()
    {
        super.onCreate();

        //register your Broadcast receiver here
        IntentFilter intentFilter = new IntentFilter("MANUAL_BROADCAST_RECIEVER");
        registerReceiver(new BroadcastReciever(), intentFilter);

    }
}

Answer 12

这就是我这样做的方式，它是Ceph3us给出并由Slinden77编辑的答案的修改版本（除其他内容外，我删除了我不需要的方法的返回值）：

public class MyBroadcastReceiver extends BroadcastReceiver{
    private boolean isRegistered; 

    public void register(final Context context) {
        if (!isRegistered){
            Log.d(this.toString(), " going to register this broadcast receiver");
            context.registerReceiver(this, new IntentFilter("MY_ACTION"));
            isRegistered = true;
        }
    }
    public void unregister(final Context context) {
        if (isRegistered) {            
            Log.d(this.toString(), " going to unregister this broadcast receiver");
            context.unregisterReceiver(this);
            isRegistered = false;
        }
    }
    @Override
    public void onReceive(final Context context, final Intent intent) {        
        switch (getResultCode()){
        //DO STUFF
        }        
    }        
}

然后在活动课上：

public class MyFragmentActivity extends SingleFragmentActivity{
    MyBroadcastReceiver myBroadcastReceiver;

    @Override
    protected void onCreate(Bundle savedInstanceState){
        super.onCreate(savedInstanceState);
        registerBroacastReceiver();       
    }

    @Override
    protected Fragment createFragment(){
        return new MyFragment();
    }

    //This method is called by the fragment which is started by this activity, 
    //when the Fragment is done, we also register the receiver here (if required)
    @Override
    public void receiveDataFromFragment(MyData data) {
        registerBroacastReceiver();
        //Do some stuff                
    }

    @Override
    protected void onStop(){        
        unregisterBroacastReceiver();
        super.onStop();
    }

    void registerBroacastReceiver(){
        if (myBroadcastReceiver == null)
            myBroadcastReceiver = new MyBroadcastReceiver();
        myBroadcastReceiver.register(this.getApplicationContext());
    }

    void unregisterReceiver(){
        if (MyBroadcastReceiver != null)
            myBroadcastReceiver.unregister(this.getApplicationContext());
    }
}

Answer 13

我将此代码放在父母活动中

列表registeredReceivers = new ArrayList <>（）;

@Override
public Intent registerReceiver(BroadcastReceiver receiver, IntentFilter filter) {
    registeredReceivers.add(System.identityHashCode(receiver));
    return super.registerReceiver(receiver, filter);
}

@Override
public void unregisterReceiver(BroadcastReceiver receiver) {
    if(registeredReceivers.contains(System.identityHashCode(receiver)))
    super.unregisterReceiver(receiver);
}

Answer 14

这是我为检查广播公司是否已经注册的工作，即使您关闭了应用程序（finish（aftery（））

运行您的应用程序的第一时间，首先发送广播，它将返回true/false取决于您的广播公司是否仍在运行中。

我的广播公司

public class NotificationReceiver extends BroadcastReceiver {
    @Override
    public void onReceive(Context context, Intent intent) {
        if(intent.getExtras() != null && intent.getStringExtra("test") != null){
            Log.d("onReceive","test");
            return;
        }
    }
}

我的主要目的

// init Broadcaster
private NotificationReceiver nr = new NotificationReceiver();


Intent msgrcv = new Intent("Msg");
msgrcv.putExtra("test", "testing");
boolean isRegistered = LocalBroadcastManager.getInstance(this).sendBroadcast(msgrcv);

if(!isRegistered){
    Toast.makeText(this,"Starting Notification Receiver...",Toast.LENGTH_LONG).show();
    LocalBroadcastManager.getInstance(this).registerReceiver(nr,new IntentFilter("Msg"));
}

Answer 15

您可以使用 匕首 创建该接收器的参考。

首先提供：

@Provides
@YourScope
fun providesReceiver(): NotificationReceiver{
    return NotificationReceiver()
}

然后将其注入您需要的地方（使用 constructor 或字段 injection)

只是将其传递给 registerReceiver.

也把它放在 try/catch 也可以阻止。

Answer 16

if( receiver.isOrderedBroadcast() ){
     // receiver object is registered
}
else{
     // receiver object is not registered
}

Answer 17

只需检查NullPoInterException即可。如果接收器不存在，那就...

try{
    Intent i = new Intent();
    i.setAction("ir.sss.smsREC");
    context.sendBroadcast(i);
    Log.i("...","broadcast sent");
}
catch (NullPointerException e)
{
    e.getMessage();
}