将像素位置从Google静态地图转换为latlng

Question

我有一个我存储在2D数组中的Google静态图。当用户单击像素时，我知道行/col [0-1024]。

我需要将该行col转换回LAT LNG。我一直在尝试使用下面的代码进行从Pixel到LAT LNG的转换。

。

我所做的是乘坐中心液化天然气和变焦级别。

1. 使用FromlatlngTopoint将中心LNG转换为像素。
2. 使用图像中的行/col以及图像的宽度和高度获取

private static PointF getPointFromRowCol(int row, int col, int width, int  height, PointF center) {
    double adjustedStartCol = center.x - ((double)width / 2);
    double adjustedStartRow = center.y - ((double)height / 2);
    double adjustedCol = adjustedStartCol + col;
    double adjustedRow = adjustedStartRow + row;
    PointF adjustedWorldPoint = new PointF(adjustedCol, adjustedRow);
    return GoogleMapsProjection2.fromPointToLatLng(adjustedWorldPoint, 17);
}

问题是，当我将生成的LNG放回Google地图中时，它被100米关闭了。

有什么想法？

public final class GoogleMapsProjection2 {
    private final int TILE_SIZE = 256;
    private PointF _pixelOrigin;
    public double _pixelsPerLonDegree;
    public double _pixelsPerLonRadian;
    public GoogleMapsProjection2() {
        this._pixelOrigin = new PointF(TILE_SIZE / 2.0, TILE_SIZE / 2.0);
        this._pixelsPerLonDegree = TILE_SIZE / 360.0;
        this._pixelsPerLonRadian = TILE_SIZE / (2 * Math.PI);
    }
    double bound(double val, double valMin, double valMax) {
        double res;
        res = Math.max(val, valMin);
        res = Math.min(val, valMax);
        return res;
    }
    double degreesToRadians(double deg) {
        return deg * (Math.PI / 180);
    }
    double radiansToDegrees(double rad) {
        return rad / (Math.PI / 180);
    }
    public PointF fromLatLngToPoint(double lat, double lng, int zoom) {
        PointF point = new PointF(0, 0);
        point.x = _pixelOrigin.x + lng * _pixelsPerLonDegree;
        // Truncating to 0.9999 effectively limits latitude to 89.189. This is
        // about a third of a tile past the edge of the world tile.
        double siny = bound(Math.sin(degreesToRadians(lat)), -0.9999, 0.9999);
        point.y = _pixelOrigin.y + 0.5 * Math.log((1 + siny) / (1 - siny)) * -_pixelsPerLonRadian;
        int numTiles = 1 << zoom;
        point.x = point.x * numTiles;
        point.y = point.y * numTiles;
        return point;
    }
    public PointF fromPointToLatLng(PointF point, int zoom) {
        int numTiles = 1 << zoom;
        point.x = point.x / numTiles;
        point.y = point.y / numTiles;
        double lng = (point.x - _pixelOrigin.x) / _pixelsPerLonDegree;
        double latRadians = (point.y - _pixelOrigin.y) / -_pixelsPerLonRadian;
        double lat = radiansToDegrees(2 * Math.atan(Math.exp(latRadians)) - Math.PI / 2);
        return new PointF(lat, lng);
    }
    public static PointF fromWorldToPixel(PointF world){
        PointF pixelPoint = new PointF(world.x * (1 << 17), world.y * (1 << 17));
        return pixelPoint;
    }
    public static PointF pixelToWorld(PointF world){
        PointF pixelPoint = new PointF(world.x / (1 << 17), world.y / (1 << 17));
        return pixelPoint;
    }
}

Answer 1

此页面似乎建议您可以将其他API与静态地图一起使用。

似乎值得一提的是，您实际上可以拥有Google Maps JavaScript API为您提供Latudinal＆amp;来自像素坐标的纵向坐标。

虽然在V3中有点令人费解，但这是一个如何做的示例。
（注意：这是假设您已经有一个地图，并且要将像素顶点转换为LAT＆amp; lng坐标）：

let overlay  = new google.maps.OverlayView();
overlay.draw = function() {};
overlay.onAdd = function() {};
overlay.onRemove = function() {};
overlay.setMap(map);
let latlngObj = overlay.fromContainerPixelToLatLng(new google.maps.Point(pixelVertex.x, pixelVertex.y);
overlay.setMap(null); //removes the overlay

希望对某人有帮助。

Answer 2

您可以尝试此Java函数：

 function convertPixelToGeo(tx:Number, ty)
{
  $LatBottomSin=min(max(sin($this->mapLatBottom*(M_PI/180)),-0.9999),0.9999);
  $worldMapWidth=(($this->mapWidth/$mapLonDelta)*360)/(2*M_PI);
  $worldMapRadius = $mapWidth / $mapLonDelta * 360/(2*M_PI);     
  $mapOffsetY = ($worldMapRadius/2 *log((1+sin($LatBottomSin))/(1-sin($LatBottomSin))));
  $equatorY = $this->mapHeight + mapOffsetY;   
  $a = ($equatorY-$ty)/$worldMapRadius;
  $lat = 180/Math.PI * (2 * Math.atan(Math.exp($a)) - Math.PI/2);
  $long = $this->mapLonLeft+$tx/$mapWidth*$mapLonDelta;
  return new Point($lat,$long);
 }

如果它不起作用，请尝试以下方法：

    int numTiles = 1 << zoom;
    tx = tx / numTiles;
    ty = ty / numTiles;