如何根据当前位置找到最近的100米纬度和经度,我已经在SQLITE3中创建了一个数据库,其中包含一组纬度和经度以及相应的位置名称。根据当前位置,我想在objective C中获取最近的100米纬度和经度。
以下是我使用的代码获取当前位置:
获取以下输出8505.369547,当返回时,我不确定实际得到的值是什么。
以下是我使用的代码获取当前位置:
- (void)viewDidLoad {
[super viewDidLoad];
databaseName = @"searchdata.sql";
NSArray *documentPaths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
NSString *documentDir = [documentPaths objectAtIndex:0];
databasePath = [documentDir stringByAppendingPathComponent:databaseName];
[self checkAndCreateDatabase];
[self readDataFromDatabase];
// Uncomment the following line to display an Edit button in the navigation bar for this view controller.
// self.navigationItem.rightBarButtonItem = self.editButtonItem;
locationManager = [[CLLocationManager alloc] init];
locationManager.delegate = self;
locationManager.distanceFilter = kCLDistanceFilterNone; // whenever we move
locationManager.desiredAccuracy = kCLLocationAccuracyHundredMeters; // 100 m
[locationManager startUpdatingLocation];
}
- (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];
NSLog(@" Current Latitude : %@",lat);
//latLabel.text = lat;
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];
NSLog(@" Current Longitude : %@",longt);
//longLabel.text = longt;
}
如何获取或计算周围100米的纬度和经度,并显示它。
感谢您的帮助,我正在使用以下公式
-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {
double dlon = convertToRadians([pLong intValue] - [longt intValue]);
double dlat = convertToRadians([pLat intValue] - [lat intValue]);
double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians([lat intValue]))) * cos(convertToRadians([pLat intValue])) * pow(sin(dlon / 2), 2);
double angle = 2 * asin(sqrt(a));
return angle * RADIO;
}
获取以下输出8505.369547,当返回时,我不确定实际得到的值是什么。