小编PMI*_*MIW的帖子

我正在学习如何使用新的Swift语言(只有Swift,没有Objective-C).要做到这一点,我想用map(MKMapView)做一个简单的视图.我想查找并更新用户的位置(例如在Apple Map应用程序中).

我试过了,但没有发生任何事:

import MapKit
import CoreLocation

class MapView : UIViewController, CLLocationManagerDelegate {

    @IBOutlet weak var map: MKMapView!
    var locationManager: CLLocationManager!

    override func viewDidLoad() {
        super.viewDidLoad()

        if (CLLocationManager.locationServicesEnabled())
        {
            locationManager = CLLocationManager()
            locationManager.delegate = self
            locationManager.desiredAccuracy = kCLLocationAccuracyBest
            locationManager.requestAlwaysAuthorization()
            locationManager.startUpdatingLocation()
        }
    }
}

请你帮助我好吗？

我有一个View.在这个观点中,我有一个Container View.在ContainerView我有一个按钮.

当我触摸ContainerView的按钮时,我想要隐藏ContainerView.

我想做那样的事情:

class ContainerView: UIViewController {

    @IBAction func closeContainerViewButton(sender: AnyObject) {
        //I try this : self.hidden = false
        //or this :    self.setVisibility(self.INVISIBLE)
    }

}

知道怎么做？

我在法国,当我尝试新func locationManager功能时,我的地图视图在旧金山找到了我.任何的想法？

override func viewDidLoad() {
    super.viewDidLoad()   

    if (CLLocationManager.locationServicesEnabled())
    {
        locationManager = CLLocationManager()
        locationManager.delegate = self
        locationManager.desiredAccuracy = kCLLocationAccuracyBest
        locationManager.distanceFilter = kCLDistanceFilterNone
        locationManager.requestAlwaysAuthorization()
        locationManager.startUpdatingLocation()
    }
}

func locationManager(manager: CLLocationManager!, didUpdateLocations locations: [AnyObject]!) {
    let location = locations[locations.count-1] as CLLocation

    let center = CLLocationCoordinate2D(latitude: location.coordinate.latitude, longitude: location.coordinate.longitude)
    let region = MKCoordinateRegion(center: center, span: MKCoordinateSpan(latitudeDelta: 0.01, longitudeDelta: 0.01))

    self.map.setRegion(region, animated: true)
}

我正在使用模拟器.