ios - CLLocation 距离跟踪开始时不正确

Question

我是 Swift 的新手(以及这个网站，如果我做错了什么，很抱歉)，我正在尝试制作一个正在运行的应用程序来跟踪用户的位置。虽然我用来跟踪距离的函数有效，但它并不是从 0 开始。当我点击开始按钮时，距离从随机数开始，然后从那里开始跟踪。

我的问题是:是否有什么我没有正确解决的问题？如果是这样，有没有办法修复它以使跟踪更加准确？这是我到目前为止所拥有的:

override func viewDidLoad() {
    super.viewDidLoad()


    stopwatchLabel.text = "00:00.00"
    locationManager.desiredAccuracy = kCLLocationAccuracyBest
    locationManager.delegate = self

    locationManager.requestAlwaysAuthorization()
    locationManager.activityType = .fitness
   locationManager.distanceFilter = 10.0
    mapView.showsUserLocation = true
    startLocation = nil

    // Do any additional setup after loading the view.
}
override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
}

// MARK: - Location Delegate Methods


func locationManager(_ manager: CLLocationManager, didUpdateLocations locations: [CLLocation])
{
    let location = locations.last
    let center = CLLocationCoordinate2D(latitude: location!.coordinate.latitude, longitude: location!.coordinate.longitude)
    let region = MKCoordinateRegion(center: center, span: MKCoordinateSpan(latitudeDelta: 0.002, longitudeDelta: 0.002))
    self.mapView.setRegion(region, animated: true)


    if startLocation == nil {
        startLocation = locations.first
    }
    var distance = startLocation.distance(from: location!)
    let lastDistance = location?.distance(from: location!)
    distance += lastDistance!
    distanceString = "\(distance)"
    distanceLabel.text = distanceString


}

该应用程序如下所示:

the run screen

我意识到其他人也问过类似的问题，但这些问题要么没有答案，要么使用不同的语言(例如 Objective-C)。如果这个问题之前已经被回答过，而我只是忽略了它，有人可以将答案链接给我吗？谢谢!

Answer 1

当位置管理器启动时，返回的第一个位置是缓存的最后知道的位置。您需要通过时间戳来检查这一点，并检查返回的准确性级别。您的 didUpdateLocations 委托(delegate)中的类似内容:

let newLocation = locations.last

    let timeDiff = newLocation?.timestamp.timeIntervalSinceNow

let accuracyNeeded:CLLocationAccuracy=100.0

    if timeDiff < 5.0 && (newLocation?.horizontalAccuracy)!<=accuracyNeeded{
        //your code here
    }