CODEWITHSUNDEEP
iosswiftiphonecore-bluetooth
Leri Gogsadze
Leri Gogsadze

Reputation: 114

How to scan non LE devices using Swift?

I have RFID reader, which is not LE device.

https://www.tsl.com/products/1153-bluetooth-wearable-uhf-rfid-reader

I'm trying to write an iOS application, scan this device and connect it using swift CoreBluetooth library but my App finds everything besides this device. How is it possible to scan this reader?

import UIKit
import CoreBluetooth

class ViewController: UIViewController, CBCentralManagerDelegate {

    var manager: CBCentralManager!

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.

        manager = CBCentralManager(delegate: self, queue: nil)
    }

    func centralManager(_ central: CBCentralManager, didDiscover peripheral: CBPeripheral, advertisementData: [String : Any], rssi RSSI: NSNumber) {
        print(peripheral)
    }

    func centralManagerDidUpdateState(_ central: CBCentralManager) {
        switch central.state {
        case .unknown:
            break;
        case .poweredOff:
            break;
        case .poweredOn:
            manager.scanForPeripherals(withServices: nil)
            break;
        case .resetting:
            break;
        case .unauthorized:
            break;
        case .unsupported:
            break;
        default:
            break;
        }
    }
}

Upvotes: 0

Views: 410

Answers (1)

Paulw11
Paulw11

Reputation: 114975

That device states that is MFi certified and uses the SPP profile, not the BLE GATT profile. This means that you will need to use the External Accessory Framework, not Core Bluetooth, to communicate with it.

You will need to the manufacturer provided iOS SDK for the device. If they do and you want to release your app on the App Store then they will also need to approve your app and supply some paperwork to Apple.

The device says that it also supports the HID profile, so perhaps you could just treat it as a keyboard; This doesn't require any code but isn't the best user experience.

Upvotes: 2

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