Android How to answer a call through handset earpiece when the headset is already connected?

Question

A headset device has be connected to my phone.Now a phone call is coming, I want to answer the call just use the telephone not by the headset. How to do this ?

Any help will be appreciate!

Answer 1

No more words,look at these codes.You should listen the PhoneStateListener's CALL_STATE_OFFHOOK and call the setHandsetPhone() at a delay time.

private void setHandsetPhone()
{
    if (isBluetoothAvailable() && isBluetoothAudioConnected())
    {
        disconnectBluetoothAudio();
    }
    //am == AudioManager
    if (am.isSpeakerphoneOn())
    {
        am.setSpeakerphoneOn(false);
    }
}

private boolean isBluetoothAvailable()
{
    printLog("isBluetoothAvailable()...");
    // Check if there's a connected headset, using the BluetoothHeadset API.
    boolean isConnected = false;
    if (mBluetoothHeadset != null)
    {
        List<BluetoothDevice> deviceList = mBluetoothHeadset.getConnectedDevices();

        if (deviceList.size() > 0)
        {
            isConnected = true;
        }
    }
    return isConnected;
}

/**
 * @return true if a BT Headset is available, and its audio is currently
 *         connected.
 */
private boolean isBluetoothAudioConnected()
{
    if (mBluetoothHeadset == null)
    {
        return false;
    }
    List<BluetoothDevice> deviceList = mBluetoothHeadset.getConnectedDevices();

    if (deviceList.isEmpty())
    {
        return false;
    }
    BluetoothDevice device = deviceList.get(0);
    boolean isAudioOn = mBluetoothHeadset.isAudioConnected(device);
    return isAudioOn;
}

private void disconnectBluetoothAudio()
{
    if (mBluetoothHeadset != null)
    {
        Class clazz = mBluetoothHeadset.getClass();
        try
        {
            Method method = clazz.getMethod("disconnectAudio",
                    new Class[] {});
            System.out.println("result-" + method.invoke(mBluetoothHeadset));
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }
    }
}