在 Swift 中使用带有静态函数的委托

Question

这是我在我的应用程序中使用的功能，效果很好。

class ResenhaEquideosMenuController: UIViewController, CLLocationManagerDelegate {

static let locationManager = CLLocationManager()

func getLocation() {
    let status = CLLocationManager.authorizationStatus()
    
    switch status {
        case .notDetermined:
                locationManager.requestWhenInUseAuthorization()
                return
            
        case .denied, .restricted:
            let alert = UIAlertController(title: "Serviços de localização desativados", message: "Por favor, ative os Serviços de Localização nas Configurações do seu dispositivo", preferredStyle: .alert)
            let okAction = UIAlertAction(title: "OK", style: .default, handler: nil)
            alert.addAction(okAction)
            
            present(alert, animated: true, completion: nil)
            return
        case .authorizedAlways, .authorizedWhenInUse:
            break
    }
    
    locationManager.delegate = self
    locationManager.startUpdatingLocation() // função didUpdateLocations controla quando GPS capta atualização do Sensor
}
}

但我想将此函数更改为静态函数，就像

static func getLocation() { // some code }

但是我在这部分代码上遇到了错误

locationManager.delegate = self

无法将类型 'ResenhaEquideosMenuController.Type' 的值分配给类型 'CLLocationManagerDelegate?'

我该如何解决？

Answer 1

静态函数不依赖于它们所属类型的任何特定实例，因此在您执行操作时从内部引用 self：

locationManager.delegate = self

没有任何意义。 self 表示为函数调用提供上下文的特定对象，静态函数不可用。

How can I fix that?

您将不得不重新考虑要使 getLocation 静态的原因，并找到不同的方法。

Answer 2

我没有找到使用具有静态函数的委托来解决我的 CCLocationManager 问题的方法 class。

因此，根据@DuncanC 的评论，我创建了一个新的 class，它扩展了一个名为 LocationViewController 的 UIViewController。

import Foundation
import CoreLocation

class LocationViewController: UIViewController, CLLocationManagerDelegate {

let locationManager = CLLocationManager()

override func viewDidLoad() {
    super.viewDidLoad()
    print("___ LocationViewController viewDidLoad")
    getLocation()
}

func getLocation() {
    let status = CLLocationManager.authorizationStatus()
    
    switch status {
        case .notDetermined:
                locationManager.requestWhenInUseAuthorization()
                return
            
        case .denied, .restricted:
            let alert = UIAlertController(title: "Serviçoõs de localização desativados", message: "Por favor, ative os Serviços de Localização nas Configurações do seu dispositivo", preferredStyle: .alert)
            let okAction = UIAlertAction(title: "OK", style: .default, handler: nil)
            alert.addAction(okAction)
            
            present(alert, animated: true, completion: nil)
            return
        case .authorizedAlways, .authorizedWhenInUse:
            break
        @unknown default:
            print("erro desconhecido")
    }
    
    locationManager.delegate = self
    locationManager.startUpdatingLocation() // função didUpdateLocations controla quando GPS capta atualização do Sensor
}

func locationManager(_ manager: CLLocationManager, didUpdateLocations locations: [CLLocation]) {
    if let currentLocation = locations.last {
        manager.distanceFilter = 50 // distance changes you want to be informed about (in meters)
        manager.desiredAccuracy = 10 // biggest approximation you tolerate (in meters)
        //manager.activityType = .automotiveNavigation // .automotiveNavigation will stop the updates when the device is not moving
        saveLocationPosition(currentLocation)
    }
}

fileprivate func saveLocationPosition(_ currentLocation: CLLocation) {
    UserDefaults.standard.set(currentLocation.coordinate.latitude, forKey: RESconstantes.LATITUDE_USER)
    UserDefaults.standard.set(currentLocation.coordinate.longitude, forKey: RESconstantes.LONGITUDE_USER)
}

func locationManager(_ manager: CLLocationManager, didFailWithError error: Error) {
    print("locationManager error")
    print(error)
}

}

当我想捕捉 GPS 传感器的位置时，我只是使用 LocateViewController 扩展了一个 UIViewController。

就像下面的例子：

class ResenhaEquideosMenuController: LocationViewController { ...code }

对我有用。

在 Swift 中使用带有静态函数的委托

Using delegate with a static function in Swift

xcode

swift