Simple Bluetooth server

I'm trying to set up a simple bluetooth server so I can ulitmately connect my sports watch to it, however I can't seem to see the server in any BLE scanning apps, making wonder whether this component is only for internal APP inventor bluetooth client <-> server applications, having re-read the docs :

Use the BluetoothServer component to turn your device into a server that receive connections from other apps that use the BluetoothClient component.

The blocks I'm using are here and the status variables are polled every second to give realtime feedback:

The app looks like this - nothing fancy at all just creating an isolated unit:

If someone can let me know if I have to resort to custom code to achieve then I'll just go down that route instead of using these components.

Thanks for any help :slight_smile:

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Thanks for the tutorial, however I can't see my BLE_A device in any BLE scanning apps. Is it only visible to MIT App inventor BLE clients?

Yes, if you want to connect your device with another device you have to use a bluetooth client, and the device you want to connect have to use the bluetooth server to accept the connection.

and I suggest you to try 2 projects below, so you can see how it works

BlueT_Server.aia (23.1 KB)
BlueT_Client.aia (17.4 KB)

Note: try two projects on different devices

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BLE is different from plain BlueTooth.
Start here ...

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Thanks for posting those articles - I've updated the extension, now to learn all the new methods :slight_smile:

So if I want to create a fake cycling power meter - how easy is that? I can connect to heart rate monitors and power meters no problem ( using standard blue tooth component within Connectivity ) however I'm not sure it's possible within an Android app to do something like this:

I'm calling the device BLE_A and giving the service UUID of Cycling Power Measurement - the hope is that my running watch will see it as a device, but suspect there's a lot more that I need to do or it's not possible ( so will have to build it on Arduino ).

Thanks everyone for their help,


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