我的 android 应用程序多次检测到同一个蓝牙设备

Question

我正在使用来自 Android 开发者网站的代码来检测范围内的蓝牙设备，并将它们添加到 ArrayAdapter。问题是，每个设备都被添加到 ArrayAdapter 5-6 次。现在，我只是使用这里的代码：http://developer.android.com/guide/topics/connectivity/bluetooth.html#DiscoveringDevices

这是我所拥有的：

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());
        }
    }
};

知道是什么原因造成的吗？我该怎么做才能让一个设备只添加到 ArrayAdapter 一次，而不是 5 次？

Answer 1

我不确定这是错误还是什么，但我在某些设备上也遇到过这种情况。为了解决这个问题，只需在List 中添加一次找到的设备并进行一些检查。见下文：

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());
                }
            }
        }
    };