Question

我创建了一个广播服务，其中包含应该等待输入的runnables，它至少运行第一次runnable内联，但不会添加任何其他的，也不会等待和监听。

public class MainService extends Service {

//  Manager device;
    Emitter emitter;

    RFIDPhidget device;

    Service service;

    @Override
    public IBinder onBind(Intent intent) {
        return null;
    }   

     @Override
     public void onStart(Intent intent, int startId) {
         // For time consuming an long tasks you can launch a new thread here...
         Toast.makeText(this, " Service Started", Toast.LENGTH_LONG).show();

     }


    /** Called when the activity is first created. */
    @Override
    public void onCreate() {
//        LinearLayout lin_out = (LinearLayout)findViewById(R.id.lineout);
//        lin_out.setVisibility(View.GONE);
        Toast.makeText(this, " Service CREATED", Toast.LENGTH_LONG).show();
        try
        {
            com.phidgets.usb.Manager.Initialize(this);

            //device = new Manager();
            device = new RFIDPhidget();

            service = this;

            emitter = new Emitter(this);

            device.addAttachListener(new AttachListener() {
                public void attached(final AttachEvent attachEvent) {
                    AttachEventHandler handler = new AttachEventHandler((RFIDPhidget)attachEvent.getSource(), service);    
                    handler.run();
                    synchronized(handler) {

                        try {
                            handler.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            });

            device.addDetachListener(new DetachListener() {
                public void detached(final DetachEvent detachEvent) {
                    DetachEventHandler handler = new DetachEventHandler((RFIDPhidget)detachEvent.getSource(), service);    
                    handler.run();
                    synchronized(handler) {
                        try {
                            handler.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            });

            device.addTagGainListener(new TagGainListener() {
                public void tagGained(TagGainEvent oe) {
                    TagGainedHandler handler = new TagGainedHandler(oe, emitter);    
                    handler.run();
                    synchronized(handler)
                    {
                        try {
                            handler.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            });

            device.addTagLossListener(new TagLossListener() {
                public void tagLost(TagLossEvent oe) {
                    TagLossHandler handler = new TagLossHandler(oe, emitter);    
                    handler.run();
                    synchronized(handler)
                    {
                        try {
                            handler.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            });


            device.openAny();
        } catch (PhidgetException pe) {
            pe.printStackTrace();
        }
    }


    @Override
    public void onDestroy() {
        super.onDestroy();
        try {
            device.close();
        } catch (PhidgetException e) {
            e.printStackTrace();
        }
        com.phidgets.usb.Manager.Uninitialize();
    }
}

我曾经在一个活动中运行这整个代码，并为每个监听器变量'handler'使用runOnUIThread(handler);但是因为这在服务中是不可能的，所以我开始使用handler.run();。

它至少运行一次AttachEventHandler，它不会运行其他的，也不会等待它。以下是侦听器内部代码的示例。

public class AttachEventHandler implements Runnable { 
    RFIDPhidget device;
    Service main_service;

    public AttachEventHandler(RFIDPhidget device, Service main_service) {
        this.device = device;
        this.main_service = main_service;
    }


    public void run() {
        try {
            device.setAntennaOn(true);
            device.setLEDOn(true);
            Toast.makeText(main_service, "Hello " + device.getDeviceName() + ", Serial " + Integer.toString(device.getSerialNumber()), Toast.LENGTH_LONG);
        } catch (PhidgetException e) {
            e.printStackTrace();
        }

        // Notify that we're done
        synchronized(this) {
            this.notify();
        }
    }

所有听众都相同，接受他们的尝试功能。

我需要帮助弄清楚如何让他们等待或至少继续在后台运行，这就是我转移到服务的全部原因所以我不会依赖于UI。

如果您需要任何进一步的信息，请不要害怕。

Answer 1

你需要调用你的处理程序，比如新的Thread（handler）.start（）;而不是handler.run（）;

因为运行，在同一个线程中执行，所以你不会得到wait / notify

Android线程不等待只运行一次

1 个答案: