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我们可以在 iPhone 的地图上设置特定位置的周边区域如下

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);

Android有没有等效的方法？

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2 回答 2

15

此代码不是生产质量。改用评论中的克里斯建议：https ://issuetracker.google.com/issues/35823607#comment4

这个问题最初是针对 Maps API v1 提出的。此答案适用于 v2，但可以轻松更改为 v1，所以...

没有简单的方法来做到这一点。

您可能想在 gmaps-api-issues 上请求此功能。

由于等待这在 Google 方面实施可能需要几个月的时间，所以这就是我要做的：

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));

笔记：

  • 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
  • 您可以删除LatLng south计算的第 3 部分，并像处理经度一样：这对于 latDistance 的小值是准确的（猜测您不会看到 100 公里以下的差异）
  • 代码很丑，所以请随意重构
于 2013-05-07T11:02:37.810 回答
3

虽然上述答案可能有效，但正如作者已经提到的那样，它并没有真正向前看。这是一些对我有用的代码。请注意，代码假设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));
于 2013-07-17T17:29:43.213 回答