Question

我希望像雷达屏幕一样，以用户所在的位置为中心，同时缩放到足以包含v2 api的目标点。现在我正在使用

bounds = new LatLngBounds.Builder().include(destPos)
                .include(yourPos).build();
        map.animateCamera(CameraUpdateFactory.newLatLngBounds(bounds, 50));

但是这个中心位于两点之间的某个点上并且缩放以包括两者。有没有像我这样容易做的事情？或者我是否需要或多或少从头开始并进行一些数学运算（例如，计算两点之间的距离，计算LatLngBounds的lat / lng，以便用户位于定义的矩形的中心，矩形的边缘包括目的地 - 考虑地图/屏幕尺寸）？

这就是我所拥有的：

enter image description here

这就是我想要的：

enter image description here

Answer 1

你自己的答案过于复杂。

您需要计算位置另一侧的点，并将其包含在LatLngBounds中。

LatLng otherSidePos = new LatLng(2 * yourPos.latitude - destPos.latitude, 2 * yourPos.longitude - destPos.longitude);
bounds = new LatLngBounds.Builder().include(destPos)
                .include(otherSidePos).build();
        map.animateCamera(CameraUpdateFactory.newLatLngBounds(bounds, padding));

注意：您不希望将50像素硬编码为填充。它在不同设备上看起来会有所不同。改为使用与密度无关的像素。

Answer 2

用于缩放以包括所有标记而不更改相机目标（地图中心）。

LatLngBounds.Builder builder = new LatLngBounds.Builder();
for (Marker marker : markers) {
    builder.include(marker.getPosition());
}
LatLngBounds bounds = builder.build();

LatLng currentLoc = _googleMapView.getCameraPosition().target;

//make sure that centre location doesn't change
double deltaLat = Math.max(Math.abs(bounds.southwest.latitude - currentLoc.latitude),
                Math.abs(bounds.northeast.latitude - currentLoc.latitude));

double deltaLon = Math.max(Math.abs(bounds.southwest.longitude - currentLoc.longitude),
                Math.abs(bounds.northeast.longitude - currentLoc.longitude));

LatLngBounds.Builder displayBuilder = new LatLngBounds.Builder();
displayBuilder.include(currentLoc); //not necessary but hey
displayBuilder.include(new LatLng(currentLoc.latitude + deltaLat, currentLoc.longitude + deltaLon));
displayBuilder.include(new LatLng(currentLoc.latitude - deltaLat, currentLoc.longitude - deltaLon));

_googleMapView.moveCamera(CameraUpdateFactory.newLatLngBounds(displayBuilder.build(), 0));

Answer 3

我想出了如何以上面给出的方式做到这一点。我为LatLngBounds设置了一个方形区域，以简化轴承编号（只要它适合屏幕，我就可以了：你可以弄清楚轴承从中心到非方形矩形的角落是什么有一些更多的trig）。我还添加了一些填充，因为点之间的距离将与从中心到东北角的对角线相同，并且这将大于从中心到顶部边缘的距离，这意味着第二个点可以在顶部边缘上方并且不可见（如果它是，例如，正北）。我从http://www.movable-type.co.uk/scripts/latlong.html和calculating Lat and Long from Bearing and Distance，

大量借用

这是获取LatLngBounds的方法：

public LatLngBounds getBounds(double lat1, double lng1, double lat2,
        double lng2) {

    // defines a square area with point lat1,lng1 at center and includes
    // point
    // lat2, long2 with a buffer between edge and second point

    // get distance between two points
    float[] results = new float[1];

    Location.distanceBetween(lat1, lng1, lat2, lng2, results);
    double d = results[0];

    d = (d * 1.5); // add padding. The shortest distance to an edge of the
//square box is d * cos(45 degrees). Thus it's possible that the second point will be out of the box. To compensate I make the radius larger (d * 1/cos(45) = d * 1.41). I put a bit extra so that the second point isn't right on the edge of the screen. 

    long R = 6371000; // distance of earth's radius in meters

    d = d /(double) R;

    lat1 = Math.toRadians(lat1); // Current lat point converted to radians
    lng1 = Math.toRadians(lng1); // Current long point converted to radians

    // calculate northeast corner of LatLngBounds

    double brng = Math.toRadians(45); // bearing from center to northeast in
                                        // radians

    double resultLat1 = Math.asin(Math.sin(lat1) * Math.cos(d)
            + Math.cos(lat1) * Math.sin(d) * Math.cos(brng));

    double resultLng1 = lng1
            + Math.atan2(Math.sin(brng) * Math.sin(d) * Math.cos(lat1),
                    Math.cos(d) - Math.sin(lat1) * Math.sin(resultLat1));

    resultLat1 = Math.toDegrees(resultLat1);
    resultLng1 = Math.toDegrees(resultLng1);

    Log.i("My Code", "resultLat1: " + resultLat1 + " resultLng1: " + resultLng1);

    LatLng northEast = new LatLng(resultLat1, resultLng1);

    // calculate southwest corner of LatLngBounds. Everything is the same
    // except the bearing value

    brng = Math.toRadians(225); // bearing from center to southwest corner
                                // in radians
    double resultLat2 = Math.asin(Math.sin(lat1) * Math.cos(d)
            + Math.cos(lat1) * Math.sin(d) * Math.cos(brng));

    double resultLng2 = lng1
            + Math.atan2(Math.sin(brng) * Math.sin(d) * Math.cos(lat1),
                    Math.cos(d) - Math.sin(lat1) * Math.sin(resultLat2));

    resultLat2 = Math.toDegrees(resultLat2);
    resultLng2 = Math.toDegrees(resultLng2);

    Log.i("My Code", "resultLat2: " + resultLat2 + " resultLng2: " + resultLng2);

    LatLng southWest = new LatLng(resultLat2, resultLng2);

    LatLngBounds bounds = new LatLngBounds(southWest, northEast);

    return bounds;
}

将Android Google中心放在一个点上，同时放大到足以包含另一个点

3 个答案: