Reputation: 31283
First off, I know this question has been asked a lot on SO. And I went through all of them trying to resolve it but to no avail.
I have a simple app that scans for Bluetooth enabled devices. Here's my code.
import CoreBluetooth
import UIKit
class ViewController: UIViewController, CBCentralManagerDelegate, CBPeripheralDelegate {
var manager: CBCentralManager!
var peripheral: CBPeripheral!
override func viewDidLoad() {
super.viewDidLoad()
manager = CBCentralManager(delegate: self, queue: nil)
}
// MARK: - CBCentralManagerDelegate
func centralManagerDidUpdateState(central: CBCentralManager) {
print(#function)
switch central.state {
case .Unsupported:
print("Unsupported")
case .Unauthorized:
print("Unauthorized")
case .PoweredOn:
print("Powered On")
central.scanForPeripheralsWithServices(nil, options: nil)
case .Resetting:
print("Resetting")
case .PoweredOff:
print("Powered Off")
case .Unknown:
print("Unknown")
}
}
func centralManager(central: CBCentralManager, didDiscoverPeripheral peripheral: CBPeripheral, advertisementData: [String : AnyObject], RSSI: NSNumber) {
print(#function)
print("Discovered \(peripheral.name) at \(RSSI)")
if peripheral.name!.containsString(name) {
manager.stopScan()
self.peripheral = peripheral
self.peripheral.delegate = self
manager.connectPeripheral(peripheral, options: nil)
}
}
func centralManager(central: CBCentralManager, didConnectPeripheral peripheral: CBPeripheral) {
print(#function)
}
func centralManager(central: CBCentralManager, didDisconnectPeripheral peripheral: CBPeripheral, error: NSError?) {
central.scanForPeripheralsWithServices(nil, options: nil)
}
// MARK: - CBPeripheralDelegate
func peripheral(peripheral: CBPeripheral, didDiscoverServices error: NSError?) {
print(#function)
guard let services = peripheral.services else {
return
}
for service in services {
print(service.UUID)
if service.UUID == serviceUUID {
peripheral.discoverCharacteristics(nil, forService: service)
}
}
}
func peripheral(peripheral: CBPeripheral, didDiscoverCharacteristicsForService service: CBService, error: NSError?) {
print(#function)
guard let characteristics = service.characteristics else {
return
}
for characteristic in characteristics {
print(characteristic.UUID)
if characteristic.UUID == scratchUUID {
peripheral.setNotifyValue(true, forCharacteristic: characteristic)
}
}
}
func peripheral(peripheral: CBPeripheral, didUpdateValueForCharacteristic characteristic: CBCharacteristic, error: NSError?) {
print(#function)
var count: UInt32 = 0
if characteristic.UUID == scratchUUID {
characteristic.value!.getBytes(&count, length: sizeof(UInt32))
print(String(format: "%llu", count))
}
}
}
I ran it in my iPhone 5 (iOS 9.3.4). I had a laptop running Windows, an Android phone, an iPad Air 2 and my Mac all Bluetooth turned on. But none of those devices were discovered. centralManagerDidUpdateState
method gets called and scanning starts but that's it. didDiscoverPeripheral
delegate method never gets called.
I repeated the process, this time ran it in the iPad Air 2 but same result.
If I go to the Bluetooth menu on the device I do see other devices being discovered.
In some SO answers, I saw that running it in the main queue would work. Like so, manager = CBCentralManager(delegate: self, queue: dispatch_get_main_queue())
. But that did not work for me either.
I'd really appreciate any help on this.
Upvotes: 3
Views: 2514
Reputation: 3093
Your other devices are probably not advertising a BLE service (like Paulw11 mentioned in the comments). All of the devices you mentioned will not appear as discoverable unless they are in the Bluetooth Settings page.
There are apps for iOS, Android, and Windows that have them advertise over LE. The easiest to use is LightBlue for iOS. Run your app on the iPhone while the other iOS device is advertising and you should see the advertisements. It might be useful to get an LE advertisement scanner on Windows or Android to double check that LightBlue is advertising.
Note Windows does not yet support Peripheral Role for Bluetooth LE, so even though you will be able to see the advertisement you will not be able to connect to it.
Upvotes: 1