Question

我正在尝试在我的应用程序中显示地图，它可以在我的模拟器和我的ipod touch 4g上运行。

但是我的iPhone总是出错，我不知道为什么会这样。也许你可以解释一下它失败的原因。错误是：

由于未捕获的异常'NSInvalidArgumentException'而终止应用，原因：'无效区域<center:nan, nan span:nan, nan>'

我已经在google和stackoverflow上搜索了解决方案，但似乎没有人遇到这个问题。

这是我的代码：

//setup mapview
mapView.delegate = self;
mapView.showsUserLocation = YES;

//set up core location
locationManager = [[CLLocationManager alloc] init];
locationManager.delegate = self;

//Set up zoom location
CLLocationCoordinate2D zoomLocation;
zoomLocation.latitude = locationManager.location.coordinate.latitude;
zoomLocation.longitude= locationManager.location.coordinate.longitude;

debug(@"Location: %f und %f",locationManager.location.coordinate.latitude,locationManager.location.coordinate.longitude);

//zoom to location
MKCoordinateRegion viewRegion = MKCoordinateRegionMakeWithDistance(zoomLocation, 0.5*METERS_PER_MILE, 0.5*METERS_PER_MILE);
MKCoordinateRegion adjustedRegion = [mapView regionThatFits:viewRegion];                
[mapView setRegion:adjustedRegion animated:YES];

正如我所说，它适用于iPod和模拟器，但不适用于iPhone。

希望你能帮助我

SOLUTION：

我太早使用了locationManager，因此应用程序崩溃了。现在我用这种方式：

      - (void)viewDidLoad {
[super viewDidLoad];

//setup mapview
mapView.delegate = self;
mapView.showsUserLocation = YES;

//set up core location
locationManager = [[CLLocationManager alloc] init];
locationManager.delegate = self;

[locationManager startUpdatingLocation];

    }

   - (void)locationManager:(CLLocationManager *)manager didUpdateToLocation:(CLLocation *)newLocation fromLocation:(CLLocation *)oldLocation{
//Set up zoom location
CLLocationCoordinate2D zoomLocation;
zoomLocation.latitude = newLocation.coordinate.latitude;
zoomLocation.longitude= newLocation.coordinate.longitude;

MKCoordinateRegion viewRegion = MKCoordinateRegionMakeWithDistance(zoomLocation, 0.5*METERS_PER_MILE, 0.5*METERS_PER_MILE);
MKCoordinateRegion adjustedRegion = [mapView regionThatFits:viewRegion];
[mapView setRegion:adjustedRegion animated:YES];

debug(@"Location: %f und %f",newLocation.coordinate.latitude,newLocation.coordinate.longitude);


    }

不知道这种方式是否100％正确，但它确实有效。

仅在一台设备上出现无效区域错误

0 个答案: