我的android应用程序多次检测到相同的蓝牙设备

Question

我正在使用安卓开发者网站上的代码来检测范围内的蓝牙设备，并将它们添加到ArrayAdapter中。问题是，每个设备都会被添加到ArrayAdapter中5-6次。现在，我只使用这里的代码：http://developer.android.com/guide/topics/connectivity/bluetooth.html#DiscoveringDevices

这就是我所拥有的：

mBluetoothAdapter = BluetoothAdapter.getDefaultAdapter();      
mBluetoothAdapter.startDiscovery();

final BroadcastReceiver mReceiver = new BroadcastReceiver() {
    public void onReceive(Context context, Intent intent) {
        String action = intent.getAction();

        // When discovery finds a device
        if (BluetoothDevice.ACTION_FOUND.equals(action)) {
            // Get the BluetoothDevice object from the Intent
            BluetoothDevice device = intent.getParcelableExtra(BluetoothDevice.EXTRA_DEVICE);

            // Add the name and address to an array adapter to show in a ListView
            mArrayAdapter.add(device.getName() + "\n" + device.getAddress());
        }
    }
};

知道这是什么原因吗？我该怎么做才能让一个设备只被添加到ArrayAdapter一次，而不是5次？

Answer 1

我不确定这是一个bug还是什么，但我在我的一些设备上也经历过这种情况。要解决此问题，只需在List中添加一次找到的设备，并进行一些检查。如下所示：

private List<BluetoothDevice> tmpBtChecker = new ArrayList<BluetoothDevice>();

    final BroadcastReceiver mReceiver = new BroadcastReceiver() {
        public void onReceive(Context context, Intent intent) {
            String action = intent.getAction();

            // When discovery starts    
            if(BluetoothAdapter.ACTION_DISCOVERY_STARTED.equals(action)){
                //clearing any existing list data
                tmpBtChecker.clear();
            }

            // When discovery finds a device
            if (BluetoothDevice.ACTION_FOUND.equals(action)) {
                // Get the BluetoothDevice object from the Intent
                BluetoothDevice device = 
                    intent.getParcelableExtra(BluetoothDevice.EXTRA_DEVICE);

                // Add the name and address to an array adapter
                if(!tmpBtChecker.contains(device)){
                   tmpBtChecker.add(device);
                   mArrayAdapter.add(device.getName()+"\n"+device.getAddress());
                }
            }
        }
    };