通过当前经度和按钮点击事件的纬度

Question

您好，我正在使用 CoreLocation 在我的地图应用程序中获取纬度和经度，通过使用以下方法获取纬度和经度

- (void)viewDidLoad {
locationManager = [[CLLocationManager alloc] init];
locationManager.delegate = self;
locationManager.distanceFilter = kCLDistanceFilterNone; // whenever we move
locationManager.desiredAccuracy = kCLLocationAccuracyHundredMeters; // 100 m
[locationManager startUpdatingLocation];
    [super viewDidLoad];
   }

_ (void)locationManager.... 方法没有得到调用

 - (void)locationManager:(CLLocationManager *)manager
didUpdateToLocation:(CLLocation *)newLocation
       fromLocation:(CLLocation *)oldLocation
  {
int degrees = newLocation.coordinate.latitude;
double decimal = fabs(newLocation.coordinate.latitude - degrees);
int minutes = decimal * 60;
double seconds = decimal * 3600 - minutes * 60;
NSString *lat = [NSString stringWithFormat:@"%d° %d' %1.4f\"", 
                 degrees, minutes, seconds];

degrees = newLocation.coordinate.longitude;
decimal = fabs(newLocation.coordinate.longitude - degrees);
minutes = decimal * 60;
seconds = decimal * 3600 - minutes * 60;
NSString *longt = [NSString stringWithFormat:@"%d° %d' %1.4f\"", 
                   degrees, minutes, seconds];

   }

我想调用 lati & 。 longi 按钮点击事件到 URL。但它不起作用。它通过 0000.0 lati &长到我的网址。有人可以告诉我如何纠正这段代码吗

Answer 1

我没有发现您的 CLLocationManager 初始化有任何明显问题。如果您没有收到回调，则可能还没有可用的位置。您还应该实现错误回调并查看它是否被调用。

我建议您尝试提供的 CLLocationManager 示例之一由 Apple 提供，看看是否可以运行该示例。

Answer 2

我这样做是为了在按钮单击时将纬度和经度传递到 Web 服务 URL...

- (void)locationManager:(CLLocationManager *)manager
didUpdateToLocation:(CLLocation *)newLocation
       fromLocation:(CLLocation *)oldLocation
  {

    if((fabs(newLocation.coordinate.latitude) > 0.001) ||    (fabs(newLocation.coordinate.longitude) > 0.001)) {
    NSLog(@"Got location %f,%f", newLocation.coordinate.latitude,   newLocation.coordinate.longitude);
    if (currentLocation != nil) {
        [currentLocation release];
    }
    currentLocation = newLocation;
    [currentLocation retain];
}}

并且在单击事件代码中我刚刚传递了 locationmanager.location.coordinate.latitude.....

NSString *url = [NSString stringWithFormat:@"http://.....Url..../hespdirectory/phpsqlsearch_genxml.php?lat=%f&lng=%f&radius=%f",locationManager.location.coordinate.latitude,locationManager.location.coordinate.longitude,radius];

Answer 3

你可以直接将 CLLocation 更改为 NSString

NSString *lat=[NSString stringWithFormat:@"%f", newLocation.latitude];
NSString *lon=[NSString stringWithFormat:@"%f", newLocation.longitude];

Answer 4

答案不会直接显示它，但您应该将控制器分配为 CLLocationManagerDelegate 才能正确接收回调？