【发布时间】:2020-01-01 01:27:01
【问题描述】:
我在我的项目中准备了以下类来检索用户当前位置:
LocationManager.swift
import Foundation
import CoreLocation
class LocationManager: NSObject {
// - Private
private let locationManager = CLLocationManager()
// - API
public var exposedLocation: CLLocation? {
return self.locationManager.location
}
override init() {
super.init()
self.locationManager.delegate = self
self.locationManager.desiredAccuracy = kCLLocationAccuracyBest
self.locationManager.requestWhenInUseAuthorization()
}
}
// MARK: - Core Location Delegate
extension LocationManager: CLLocationManagerDelegate {
func locationManager(_ manager: CLLocationManager,
didChangeAuthorization status: CLAuthorizationStatus) {
switch status {
case .notDetermined : print("notDetermined") // location permission not asked for yet
case .authorizedWhenInUse : print("authorizedWhenInUse") // location authorized
case .authorizedAlways : print("authorizedAlways") // location authorized
case .restricted : print("restricted") // TODO: handle
case .denied : print("denied") // TODO: handle
default : print("unknown") // TODO: handle
}
}
}
// MARK: - Get Placemark
extension LocationManager {
func getPlace(for location: CLLocation,
completion: @escaping (CLPlacemark?) -> Void) {
let geocoder = CLGeocoder()
geocoder.reverseGeocodeLocation(location) { placemarks, error in
guard error == nil else {
print("*** Error in \(#function): \ (error!.localizedDescription)")
completion(nil)
return
}
guard let placemark = placemarks?[0] else {
print("*** Error in \(#function): placemark is nil")
completion(nil)
return
}
completion(placemark)
}
}
}
但我不确定如何在使用 SwiftUI 时从我的 ContentView 文件中使用它。 如果不使用在标准 ViewController 中使用的方法,我应该如何获得暴露位置(在这种情况下,使用 guard、let 和 return 当然会产生各种错误,因为我不应该在这个上下文,如果我做对了)。 关于如何实现这一目标的任何提示? 我想在按下按钮时获取用户位置(目前我只使用了模型数据)。
ContentView.swift
import SwiftUI
struct Location: Identifiable {
// When conforming to the protocol Identifiable we have to to implement a variable called id however this variable does not have to be an Int. The protocol only requires that the type of the variable id is actually Hashable.
// Note: Int, Double, String and a lot more types are Hashable
let id: Int
let country: String
let state: String
let town: String
}
struct ContentView: View {
// let’s make our variable a @State variable so that as soon as we change its value (by for eexample adding new elements) our view updates automagically.
@State var locationList = [
Location(id: 0, country: "Italy", state: "", town: "Finale Emilia"),
Location(id: 1, country: "Italy", state: "", town: "Bologna"),
Location(id: 2, country: "Italy", state: "", town: "Modena"),
Location(id: 3, country: "Italy", state: "", town: "Reggio Emilia"),
Location(id: 4, country: "USA", state: "CA", town: "Los Angeles")
]
// - Constants
private let locationManager = LocationManager()
// THIS IS NOT POSSIBLE WITH SWIFTUI AND GENERATES ERRORS
guard let exposedLocation = self.locationManager.exposedLocation else {
print("*** Error in \(#function): exposedLocation is nil")
return
}
var body: some View {
// Whenever we use a List based of an Array we have to let the List know how to identify each row as unique
// When confirming to the Identifiable protocol we no longer have to explicitly tell the List how the elements in our Array (which are conforming to that protocol) are uniquely identified
NavigationView {
// let’s add a title to our Navigation view and make sure you always do so on the first child view inside of your Navigation view
List(locationList) { location in
NavigationLink(destination: LocationDetail(location: location)) {
HStack {
Text(location.country)
Text(location.town).foregroundColor(.blue)
}
}
}
.navigationBarTitle(Text("Location"))
.navigationBarItems(
trailing: Button(action: addLocation, label: { Text("Add") }))
}
}
func addLocation() {
// We are using the native .randomElement() function of an Array to get a random element. The returned element however is optional. That is because in the case of the Array being empty that function would return nil. That’s why we append the returned value only in the case it doesn’t return nil.
if let randomLocation = locationList.randomElement() {
locationList.append(randomLocation)
}
}
}
struct ContentView_Previews: PreviewProvider {
static var previews: some View {
ContentView()
}
}
【问题讨论】:
-
我添加了一个关于如何检索暴露位置的问题的回复。无论如何,在您的代码中,您忘记添加
startUpdatingLocation和获取新位置的相关方法。无论如何,我还没有实现 getPlace 方法,如果您对此有疑问,请询问(或者最好打开一个新问题)
标签: ios swiftui core-location cllocationmanager