1

我有 CLLocation 对象,其中包含用户的当前位置,并且对于可以倾斜的矩形的每个角,我有 4 个纬度/经度对。现在我想检查 CLLocation 坐标是否在该矩形内。

以下是矩形的坐标

#define NorthEast_LAT 51.514894
#define NorthEast_LNG -0.135306
#define SouthEast_LAT 51.514831
#define SouthEast_LNG -0.135153
#define NorthWest_LAT 51.514719
#define NorthWest_LNG -0.135858
#define SouthWest_LAT 51.514556
#define SouthWest_LNG -0.135714

我尝试了以下代码,但我认为它仅在矩形角度为 0 度时才有效。

 BOOL withinRect = [delegate.CLController latlngWithInBox:location 
                                                    point1:CLLocationCoordinate2DMake(NorthEast_LAT, NorthEast_LNG) 
                                                    point2:CLLocationCoordinate2DMake(SouthEast_LAT, SouthEast_LNG) 
                                                    point3:CLLocationCoordinate2DMake(NorthWest_LAT, NorthWest_LNG)
                                                    point4:CLLocationCoordinate2DMake(SouthWest_LAT, SouthWest_LNG)];

 - (BOOL) latlngWithInBox:(CLLocation *)position point1:(CLLocationCoordinate2D)point1 point2:(CLLocationCoordinate2D)point2 point3:(CLLocationCoordinate2D)point3 point4:(CLLocationCoordinate2D)point4 {

if (position.coordinate.latitude >= point3.latitude && position.coordinate.latitude <= point2.latitude
    && position.coordinate.longitude >= point3.longitude && position.coordinate.longitude <= point2.longitude) {
    return YES;
}
return NO;

}

4

3 回答 3

8

确定某个点是否在地图矩形内的另一种方法是使用MKMapKit's 函数:

  1. MKMapPointForCoordinate- 将坐标转换为地图点
  2. MKMapRectMake- 使用这些点创建矩形
  3. MKMapRectContainsPoint- 确定指定的地图点是否位于矩形内

优点是 MKMapKit (MKMapPoint, MKMapRect) 使用地图的墨卡托投影,所以不需要提供球体计算。但其中一些功能在 iOS 4.0 及更高版本中可用。

更新:

// 1 ------- 2
// |         |
// |      x  |
// |         |
// 3 ------- 4    

// 1 = topLeftCorner
// 4 = bottomRightCorner
// x = targetCoordinate

CLLocationCoordinate2D topLeftCorner = /* some coordinate */, bottomRightCorner = /* some coordinate */;
CLLocationCoordinate2D targetCoordinate = /* some coordinate */;
MKMapPoint topLeftPoint = MKMapPointForCoordinate(topLeftCorner);
MKMapPoint bottomRightPoint = MKMapPointForCoordinate(bottomRightCorner);
MKMapRect mapRect = MKMapRectMake(topLeftPoint.x, topLeftPoint.y, bottomRightPoint.x - topLeftPoint.x, bottomRightPoint.y - topLeftPoint.y);
MKMapPoint targetPoint = MKMapPointForCoordinate(targetCoordinate);   

BOOL isInside = MKMapRectContainsPoint(mapRect, targetPoint);
于 2012-05-18T13:35:25.767 回答
1

让我们忽略地理坐标的矩形在椭球体上(那里的数学很困难)。所以你想知道一个点是否在一个四边形内。

最简单的方法是首先添加一个限制:必须按照一定的顺序(NE、NW、SE、SW)给出点。然后,将它们视为正常的 2D 坐标(经度 = x,纬度 = y)。

下一步是减少问题:让坐标形成 2 个三角形,例如 NE-NW-SE 和 NW-SE-SW。然后检查您的点是否在这两个三角形之一内

于 2012-05-18T12:48:46.347 回答
0
- (BOOL) latlngWithInBox:(CLLocation *)position point1:(CLLocationCoordinate2D)point1 point2:(CLLocationCoordinate2D)point2 point3:(CLLocationCoordinate2D)point3 point4:(CLLocationCoordinate2D)point4 {
    //&& position.coordinate.latitude >= [[point4 objectAtIndex:0] floatValue] && position.coordinate.latitude <= [[point1 objectAtIndex:0] floatValue] && position.coordinate.longitude <= [[point4 objectAtIndex:1] floatValue] && position.coordinate.longitude >= [[point1 objectAtIndex:1] floatValue]
    if (PointInTriangle(position.coordinate, point1, point2, point3) || PointInTriangle(position.coordinate, point2, point3, point4)) {
        return YES;
    }
    return NO;
}

float sign(CLLocationCoordinate2D p1, CLLocationCoordinate2D p2, CLLocationCoordinate2D p3)
{
    return (p1.longitude - p3.longitude) * (p2.latitude - p3.latitude) - (p2.longitude - p3.longitude) * (p1.latitude - p3.latitude);
}

bool PointInTriangle(CLLocationCoordinate2D pt, CLLocationCoordinate2D v1, CLLocationCoordinate2D v2, CLLocationCoordinate2D v3)
{
    bool b1, b2, b3;

    b1 = sign(pt, v1, v2) < 0.0f;
    b2 = sign(pt, v2, v3) < 0.0f;
    b3 = sign(pt, v3, v1) < 0.0f;

//    NSLog(@"b1-%@", [NSNumber numberWithBool:b1]);
//    NSLog(@"b2-%@", [NSNumber numberWithBool:b2]);
//    NSLog(@"b3-%@", [NSNumber numberWithBool:b3]);

    return ((b1 == b2) && (b2 == b3));
}
于 2012-05-18T13:49:16.673 回答