4

我正在使用 GoogleMaps SDK，目前我正在尝试将 GMSVisibleRegion 转换为 CLRegion。

GMSVisibleRegion 定义为：

typedef struct {
  CLLocationCoordinate2D nearLeft;
  CLLocationCoordinate2D nearRight;
  CLLocationCoordinate2D farLeft;
  CLLocationCoordinate2D farRight;
} GMSVisibleRegion;

最快的方法是什么？

不幸的是，很难理解开发人员命名“近”和“远”的含义。我认为这条评论也很有用：

/**
 * Returns the region (four location coordinates) that is visible according to
 * the projection.
 *
 * The visible region can be non-rectangular. The result is undefined if the
 * projection includes points that do not map to anywhere on the map (e.g.,
 * camera sees outer space).
 */
 - (GMSVisibleRegion)visibleRegion;

非常感谢！

编辑：好的，我的第一步是创建一个 GMSVisibleRegion 的 MKCoordinateRegion。

我建议使用以下代码将 aa GMSVisibleRegion 转换为 MKCoordinateRegion。任何异议。


+ (MKCoordinateRegion)regionForCenter:(CLLocationCoordinate2D)center andGMSVisibleRegion:(GMSVisibleRegion)visibleRegion
{
    CLLocationDegrees latitudeDelta = visibleRegion.farLeft.latitude - visibleRegion.nearLeft.latitude;
    CLLocationDegrees longitudeDelta = visibleRegion.farRight.longitude - visibleRegion.farLeft.longitude;
    MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);

    return MKCoordinateRegionMake(center, span);
}
4

4 回答 4

11

我的猜测是“近”是指屏幕底部视图的角落，而“远”是指屏幕顶部的角落。这是因为如果您倾斜了视图，那么底角离相机最近，顶角离相机最远。

将其转换为 a 的一种方法CLRegion可能是使用相机的目标作为中心，然后计算从最大距离到四个角的半径。这可能不是该区域上最紧密的拟合圆，但由于圆无论如何都无法拟合视图的四边形，因此它可能足够接近。

这是一个辅助函数，用于计算两个CLLocationCoordinate值之间的距离（以米为单位）：

double getDistanceMetresBetweenLocationCoordinates(
    CLLocationCoordinate2D coord1, 
    CLLocationCoordinate2D coord2)
{
    CLLocation* location1 = 
        [[CLLocation alloc] 
            initWithLatitude: coord1.latitude 
            longitude: coord1.longitude];
    CLLocation* location2 = 
        [[CLLocation alloc] 
            initWithLatitude: coord2.latitude 
            longitude: coord2.longitude];

    return [location1 distanceFromLocation: location2];
}

然后CLRegion可以这样计算：

GMSMapView* mapView = ...;
...
CLLocationCoordinate2D centre = mapView.camera.target;
GMSVisibleRegion* visibleRegion = mapView.projection.visibleRegion;

double nearLeftDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.nearLeft);
double nearRightDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.nearRight);
double farLeftDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.farLeft);
double farRightDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.farRight);
double radiusMetres = 
    MAX(nearLeftDistanceMetres, 
    MAX(nearRightDistanceMetres, 
    MAX(farLeftDistanceMetres, farRightDistanceMetres)));

CLRegion region = [[CLRegion alloc] 
    initCircularRegionWithCenter: centre radius: radius identifier: @"id"];

更新：

关于您的更新MKCoordinateRegion，您的示例代码可能不起作用。如果地图已旋转 90 度，则farLeft和nearLeft将具有相同的纬度，farRight并且farLeft将具有相同的经度，因此您的纬度和经度增量为零。

您需要遍历所有四个farLeft, farRight, nearLeft, nearRight，计算每个的纬度和经度的最小值和最大值，然后从中计算增量。

适用于 iOS 的 Google Maps SDK 包含一个帮助程序类，它已经为您完成了一些工作 - GMSCoordinateBounds. 它可以用 a 初始化GMSVisibleRegion：

GMSMapView* mapView = ...;
....
GMSVisibleRegion visibleRegion = mapView.projection.visibleRegion;
GMSCoordinateBounds bounds = 
    [[GMSCoordinateBounds alloc] initWithRegion: visibleRegion];

然后GMSCoordinateBounds有northEast和southWest定义边界的属性。因此，您可以按如下方式计算增量：

CLLocationDegrees latitudeDelta = 
    bounds.northEast.latitude - bounds.southWest.latitude;
CLLocationDegrees longitudeDelta = 
    bounds.northEast.longitude - bounds.southWest.longitude;

您还可以从边界计算中心，因此MKCoordinateRegion：

CLLocationCoordinate2D centre = CLLocationCoordinate2DMake(
    (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
    (bounds.southWest.longitude + bounds.northEast.longitude) / 2);
MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);
return MKCoordinateRegionMake(centre, span);
于 2013-03-05T00:25:09.347 回答
9

纯粹主义者的附录

如果您想绝对严格，则需要围绕国际日期变更线进行更正。在大多数应用程序中这将是浪费精力，但这个问题最近一直让我感到非常悲痛，所以我想把它扔进社区帽子里

基于 Druce 的更新（恐怕我无法发表评论）......

GMSMapView* mapView = ...;
....
GMSVisibleRegion visibleRegion = mapView.projection.visibleRegion;
GMSCoordinateBounds bounds = 
    [[GMSCoordinateBounds alloc] initWithRegion: visibleRegion];

纬度不需要做任何事情

CLLocationDegrees latitudeDelta = 
bounds.northEast.latitude - bounds.southWest.latitude;

该协议认为，跨越国际日期变更线的地区可能在日本有一个西南角（+140 经度），在阿拉斯加有一个东北角（-150 经度）。加起来除以 2 在地球的错误一侧给出了一个点。

northEast.longitude 小于 southWest.longitude 的特殊情况需要处理

CLLocationCoordinate2D centre;
CLLocationDegrees longitudeDelta;

if(bounds.northEast.longitude >= bounds.southWest.longitude) {
//Standard case
    centre = CLLocationCoordinate2DMake(
             (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
             (bounds.southWest.longitude + bounds.northEast.longitude) / 2);
    longitudeDelta = bounds.northEast.longitude - bounds.southWest.longitude;
} else {
//Region spans the international dateline
    centre = CLLocationCoordinate2DMake(
             (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
             (bounds.southWest.longitude + bounds.northEast.longitude + 360) / 2);
    longitudeDelta = bounds.northEast.longitude + 360 
                    - bounds.southWest.longitude;
}
MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);
return MKCoordinateRegionMake(centre, span);
于 2013-05-16T12:40:58.763 回答
4

对于根据迄今为止提供的所有答案和更正寻找样板代码的任何人，这里是region作为 GMSMapView 上的一个类别实现的：

//
//  GMSMapViewExtensions.h
//

#import <Foundation/Foundation.h>
#import <MapKit/MapKit.h>
#import <GoogleMaps/GoogleMaps.h>

@interface GMSMapView (GMSMapViewExtensions)

@end

和

//
//  GMSMapViewExtensions.m
//

#import "GMSMapViewExtensions.h"

@implementation GMSMapView (GMSMapViewExtensions)

- (MKCoordinateRegion) region {
    GMSVisibleRegion visibleRegion = self.projection.visibleRegion;
    GMSCoordinateBounds * bounds = [[GMSCoordinateBounds alloc] initWithRegion: visibleRegion];

    CLLocationDegrees latitudeDelta = bounds.northEast.latitude - bounds.southWest.latitude;

    CLLocationCoordinate2D centre;
    CLLocationDegrees longitudeDelta;

    if (bounds.northEast.longitude >= bounds.southWest.longitude) {
        // Standard case
        centre = CLLocationCoordinate2DMake(
            (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
            (bounds.southWest.longitude + bounds.northEast.longitude) / 2);
        longitudeDelta = bounds.northEast.longitude - bounds.southWest.longitude;
    } else {
        // Region spans the international dateline
        centre = CLLocationCoordinate2DMake(
            (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
            (bounds.southWest.longitude + bounds.northEast.longitude + 360) / 2);
        longitudeDelta = bounds.northEast.longitude + 360 - bounds.southWest.longitude;
    }

    MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);
    return MKCoordinateRegionMake(centre, span);
}


- (MKMapRect)visibleMapRect {
    MKCoordinateRegion region = [self region];
    MKMapPoint a = MKMapPointForCoordinate(CLLocationCoordinate2DMake(
        region.center.latitude + region.span.latitudeDelta / 2,
        region.center.longitude - region.span.longitudeDelta / 2));
     MKMapPoint b = MKMapPointForCoordinate(CLLocationCoordinate2DMake(
        region.center.latitude - region.span.latitudeDelta / 2,
        region.center.longitude + region.span.longitudeDelta / 2));
     return MKMapRectMake(MIN(a.x, b.x), MIN(a.y, b.y), ABS(a.x - b.x), ABS(a.y - b.y));
}

@end

使用示例：

GMSMapView * mapView = .... // init code
MKCoordinateRegion mapRegion = mapView.region;
于 2014-05-22T13:31:07.930 回答
4

根据@Saxon Druce 的回答，这是设置和region使用 MKCoordinateRegion的快速扩展GMSMapView

extension GMSMapView {
    var region : MKCoordinateRegion {
        get {
            let position = self.camera
            let visibleRegion = self.projection.visibleRegion()
            let bounds = GMSCoordinateBounds(region: visibleRegion)
            let latitudeDelta = bounds.northEast.latitude - bounds.southWest.latitude
            let longitudeDelta = bounds.northEast.longitude - bounds.southWest.longitude
            let center = CLLocationCoordinate2DMake(
                (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
                (bounds.southWest.longitude + bounds.northEast.longitude) / 2)
            let span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta)
            return MKCoordinateRegionMake(center, span)
        }
        set {
            let northEast = CLLocationCoordinate2DMake(newValue.center.latitude - newValue.span.latitudeDelta/2, newValue.center.longitude - newValue.span.longitudeDelta/2)
            let southWest = CLLocationCoordinate2DMake(newValue.center.latitude + newValue.span.latitudeDelta/2, newValue.center.longitude + newValue.span.longitudeDelta/2)
            let bounds = GMSCoordinateBounds(coordinate: northEast, coordinate: southWest)
            let update = GMSCameraUpdate.fitBounds(bounds, withPadding: 0)
            self.moveCamera(update)
        }
    }
}
于 2015-06-19T12:32:11.277 回答