Question

我们可以在iPhone上设置地图上特定位置的周边区域，如下所示

CLLocationCoordinate2D coord = {latitude:37.09024, longitude:-95.712891};
CLLocationDistance latitudinalMeters;
latitudinalMeters =NoOfMiles * 1609.344;
CLLocationDistance longitudinalMeters;
longitudinalMeters = NoOfMiles * 1609.344;
mapViewHome.region = MKCoordinateRegionMakeWithDistance(coord, latitudinalMeters, longitudinalMeters);

是否有适用于Android的等效方法？

Answer 1

此代码不生产质量。请在此处使用来自评论的Chris建议：https://issuetracker.google.com/issues/35823607#comment4

此问题最初是针对Maps API v1提出的。这个答案适用于v2，但可以很容易地改为v1，所以......

没有简单的方法可以做到。

您可能需要request this feature on gmaps-api-issues。

等待在Google端实施此操作可能需要几个月的时间，所以这就是我要做的事情：

private static final double ASSUMED_INIT_LATLNG_DIFF = 1.0;
private static final float ACCURACY = 0.01f;

public static LatLngBounds boundsWithCenterAndLatLngDistance(LatLng center, float latDistanceInMeters, float lngDistanceInMeters) {
    latDistanceInMeters /= 2;
    lngDistanceInMeters /= 2;
    LatLngBounds.Builder builder = LatLngBounds.builder();
    float[] distance = new float[1];
    {
        boolean foundMax = false;
        double foundMinLngDiff = 0;
        double assumedLngDiff = ASSUMED_INIT_LATLNG_DIFF;
        do {
            Location.distanceBetween(center.latitude, center.longitude, center.latitude, center.longitude + assumedLngDiff, distance);
            float distanceDiff = distance[0] - lngDistanceInMeters;
            if (distanceDiff < 0) {
                if (!foundMax) {
                    foundMinLngDiff = assumedLngDiff;
                    assumedLngDiff *= 2;
                } else {
                    double tmp = assumedLngDiff;
                    assumedLngDiff += (assumedLngDiff - foundMinLngDiff) / 2;
                    foundMinLngDiff = tmp;
                }
            } else {
                assumedLngDiff -= (assumedLngDiff - foundMinLngDiff) / 2;
                foundMax = true;
            }
        } while (Math.abs(distance[0] - lngDistanceInMeters) > lngDistanceInMeters * ACCURACY);
        LatLng east = new LatLng(center.latitude, center.longitude + assumedLngDiff);
        builder.include(east);
        LatLng west = new LatLng(center.latitude, center.longitude - assumedLngDiff);
        builder.include(west);
    }
    {
        boolean foundMax = false;
        double foundMinLatDiff = 0;
        double assumedLatDiffNorth = ASSUMED_INIT_LATLNG_DIFF;
        do {
            Location.distanceBetween(center.latitude, center.longitude, center.latitude + assumedLatDiffNorth, center.longitude, distance);
            float distanceDiff = distance[0] - latDistanceInMeters;
            if (distanceDiff < 0) {
                if (!foundMax) {
                    foundMinLatDiff = assumedLatDiffNorth;
                    assumedLatDiffNorth *= 2;
                } else {
                    double tmp = assumedLatDiffNorth;
                    assumedLatDiffNorth += (assumedLatDiffNorth - foundMinLatDiff) / 2;
                    foundMinLatDiff = tmp;
                }
            } else {
                assumedLatDiffNorth -= (assumedLatDiffNorth - foundMinLatDiff) / 2;
                foundMax = true;
            }
        } while (Math.abs(distance[0] - latDistanceInMeters) > latDistanceInMeters * ACCURACY);
        LatLng north = new LatLng(center.latitude + assumedLatDiffNorth, center.longitude);
        builder.include(north);
    }
    {
        boolean foundMax = false;
        double foundMinLatDiff = 0;
        double assumedLatDiffSouth = ASSUMED_INIT_LATLNG_DIFF;
        do {
            Location.distanceBetween(center.latitude, center.longitude, center.latitude - assumedLatDiffSouth, center.longitude, distance);
            float distanceDiff = distance[0] - latDistanceInMeters;
            if (distanceDiff < 0) {
                if (!foundMax) {
                    foundMinLatDiff = assumedLatDiffSouth;
                    assumedLatDiffSouth *= 2;
                } else {
                    double tmp = assumedLatDiffSouth;
                    assumedLatDiffSouth += (assumedLatDiffSouth - foundMinLatDiff) / 2;
                    foundMinLatDiff = tmp;
                }
            } else {
                assumedLatDiffSouth -= (assumedLatDiffSouth - foundMinLatDiff) / 2;
                foundMax = true;
            }
        } while (Math.abs(distance[0] - latDistanceInMeters) > latDistanceInMeters * ACCURACY);
        LatLng south = new LatLng(center.latitude - assumedLatDiffSouth, center.longitude);
        builder.include(south);
    }
    return builder.build();
}

用法：

LatLngBounds bounds = AndroidMapsExtensionsUtils.boundsWithCenterAndLatLngDistance(new LatLng(51.0, 19.0), 1000, 2000);
map.moveCamera(CameraUpdateFactory.newLatLngBounds(bounds, 0));

注意：

此代码尚未经过全面测试，可能不适用于边缘情况
您可能需要调整私有常量以使其执行得更快
你可以删除计算LatLng south的第3部分，并对经度进行处理：这对于latDistance的小值是准确的（猜测你不会看到100公里以下的差异）
代码很难看，所以随意重构

Answer 2

虽然上述答案可能有效，但正如作者已经提到的那样，它并没有真正直截了当。这是一些适合我的代码。请注意，代码假定地球是一个完美的球体。

double latspan = (latMeters/111325);
double longspan = (longMeters/111325)*(1/ Math.cos(Math.toRadians(location.latitude))); 

LatLngBounds bounds = new LatLngBounds(
    new LatLng(location.latitude-latspan, location.longitude-longspan),
    new LatLng(location.latitude+latspan, location.longitude+longspan));

MKCoordinateRegionMakeWithDistance等效于Android

2 个答案: