[Interest] Detect connected Bluetooth keyboard under Android

Roman Wüger roman.wueger at gmx.at
Thu Nov 1 10:58:21 CET 2018


Hi Alex,
thanks for the hint. My plan was that the user should pair and connect the keyboard manually in the settings dialog. In the application i only want to know if a keyboard is connected, otherwise I will show an onscreen keyboard.

Thanks

Regards
Roman


> Am 30.10.2018 um 08:24 schrieb Alex Blasche <alexander.blasche at qt.io>:
> 
> As a disclaimer I have not really done any HID work on Android, but based on the Android docs you need to use the BluetoothHidDevice class to interact with such keyboards. BluetoothHidDevice.Callback.onConnectionStateChanged is probably what you need.
> 
> This Android API is not available via Qt APIs. QBluetoothLocalDevice is for the local Bluetooth device which is not the remote Bluetooth keyboard and HID connects/disconnects cannot be done programmatically by design. The user has to make the connection manually. Therefore I am assuming the platform does not expose such devices via the usual discovery mechanisms either. You have to write your own Java/JNI code to interact with the class.
> 
> --
> Alex
> 
> ________________________________________
> From: Interest <interest-bounces+alexander.blasche=qt.io at qt-project.org> on behalf of Roman Wüger <roman.wueger at gmx.at>
> Sent: Monday, 29 October 2018 11:37:32 PM
> To: Qt Project MailingList
> Subject: [Interest] Detect connected Bluetooth keyboard under Android
> 
> Hi @all,
> 
> i played with the QBluetoothLocalDevice and QBluetoothDiscoveryAgent to detected if a Bluetooth keyboard is connected to an android tablet.
> 
> However, I can’t find out the correct way. I tried to read the majordeviceclass and minordeviceclass but it doesn‘t work, I only get 31,0 but the documentation says that 5 should be the major device class. The keyboard works in any input fields, but I can‘t find it.
> 
> I only need a signal if a keyboard connects/disconnects like hardwareKeyboardConnected(bool)
> 
> What do I need for that?
> 
> Thanks in advance
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