我在Unity应用程序中获取Google静态地图的界限时出现问题。我已尝试在stackoverflow上找到多个答案。
我试图计算角落的地图是this一个。
首先，我尝试将marcelo的代码从javascript翻译为C＃。这被翻译成下面的课程。
CREATE TABLE d_dim (id int, flag varchar2(4));
INSERT ALL INTO d_dim (id, flag) VALUES (1, NULL)
INTO d_dim (id, flag) VALUES (2, NULL)
INTO d_dim (id, flag) VALUES (3, NULL)
INTO d_dim (id, flag) VALUES (4, NULL)
SELECT * FROM dual;
CREATE TABLE t_temp (id int)
;
INSERT ALL
INTO t_temp (id) VALUES (1)
INTO t_temp (id) VALUES (3)
SELECT * FROM dual;
所以起初我只尝试了地图的西南角并且它非常接近，但它有点偏离。
using System;
using UnityEngine;
public class MercatorProjection
{
static double MERCATOR_RANGE = 256;
Point pixelsOrigin;
double pixelsPerLonDegree;
double pixelsPerLonRadian;
public MercatorProjection()
{
pixelsOrigin = new Point(MERCATOR_RANGE / 2, MERCATOR_RANGE / 2);
pixelsPerLonDegree = MERCATOR_RANGE / 360;
pixelsPerLonRadian = MERCATOR_RANGE / (2 * Math.PI);
}
double bound(double value, double opt_min, double opt_max)
{
if (opt_min != 0)
value = Math.Max(value, opt_min);
if (opt_max != 0)
value = Math.Min(value, opt_max);
return value;
}
double degreesToRadians(double deg)
{
return deg * (Math.PI / 180);
}
double radiansToDegrees(double rad)
{
return rad / (Math.PI / 180);
}
Point fromLatLonToPoint(Coordinates latLon)
{
Point point = new Point();
Point origin = pixelsOrigin;
point.X = origin.X + latLon.Longitude * pixelsPerLonDegree;
double sinY = bound(Math.Sin(degreesToRadians(latLon.Latitude)), -0.9999, 0.9999);
point.Y = origin.Y + 0.5 * Math.Log((1 + sinY) / (1 - sinY)) * -pixelsPerLonRadian;
return point;
}
Coordinates fromPointToLatlon(Point point)
{
Point origin = pixelsOrigin;
Coordinates latLon = new Coordinates();
latLon.Latitude = (point.X - origin.X) / pixelsPerLonDegree;
double latRadians = (point.Y - origin.Y) / -pixelsPerLonRadian;
latLon.Longitude = radiansToDegrees(2 * Math.Atan(Math.Exp(latRadians)) - Math.PI / 2);
return latLon;
}
public void GetCorners(Coordinates center, float zoom, float mapWidth, float mapHeight)
{
double scale = Math.Pow(2, zoom);
Point centerPx = fromLatLonToPoint(center);
Point SWPoint = new Point(centerPx.X - (mapWidth / 2) / scale, centerPx.Y + (mapHeight / 2) / scale);
Coordinates SWLatLon = fromPointToLatlon(SWPoint);
Debug.Log(SWLatLon.Latitude + " " + SWLatLon.Longitude + " " + SWPoint.X + " " + SWPoint.Y);
}
}
带来this结果。
由于有很多人投票，所以我尝试翻译marcelo他的答案，我认为在将代码翻译成C＃时我做错了。
我偶然发现peterjb's answer已经完成了对C＃的javascript答案的翻译，甚至举例说明了他的代码获得了Google静态地图的正确界限。因此，在下面的类中尝试peterjb代码之后;
MercatorProjection.GetCorners(new Coordinates(Coordinates.Longitude, Coordinates.Latitude), 16, 640, 640);
我尝试通过调用以下
再次计算SW和NE角using System;
public static class GoogleMapsAPI{
static GoogleMapsAPI()
{
OriginX = TileSize / 2;
OriginY = TileSize / 2;
PixelsPerLonDegree = TileSize / 360.0;
PixelsPerLonRadian = TileSize / (2 * Math.PI);
}
public static int TileSize = 256;
public static double OriginX, OriginY;
public static double PixelsPerLonDegree;
public static double PixelsPerLonRadian;
public static double DegreesToRadians(double deg)
{
return deg * Math.PI / 180.0;
}
public static double RadiansToDegrees(double rads)
{
return rads * 180.0 / Math.PI;
}
public static double Bound(double value, double min, double max)
{
value = Math.Min(value, max);
return Math.Max(value, min);
}
//From Lat, Lon to World Coordinate X, Y. I'm being explicit in assigning to
//X and Y properties.
public static Coordinate Mercator(double latitude, double longitude)
{
double siny = Bound(Math.Sin(DegreesToRadians(latitude)), -.9999, .9999);
Coordinate c = new Coordinate(0, 0);
c.X = OriginX + longitude * PixelsPerLonDegree;
c.Y = OriginY + .5 * Math.Log((1 + siny) / (1 - siny)) * -PixelsPerLonRadian;
return c;
}
//From World Coordinate X, Y to Lat, Lon. I'm being explicit in assigning to
//Latitude and Longitude properties.
public static Coordinate InverseMercator(double x, double y)
{
Coordinate c = new Coordinate(0, 0);
c.Longitude = (x - OriginX) / PixelsPerLonDegree;
double latRadians = (y - OriginY) / -PixelsPerLonRadian;
c.Latitude = RadiansToDegrees(Math.Atan(Math.Sinh(latRadians)));
return c;
}
public static MapCoordinates GetBounds(Coordinate center, int zoom, int mapWidth, int mapHeight)
{
var scale = Math.Pow(2, zoom);
var centerWorld = Mercator(center.Latitude, center.Longitude);
var centerPixel = new Coordinate(0, 0);
centerPixel.X = centerWorld.X * scale;
centerPixel.Y = centerWorld.Y * scale;
var NEPixel = new Coordinate(0, 0);
NEPixel.X = centerPixel.X + mapWidth / 2.0;
NEPixel.Y = centerPixel.Y - mapHeight / 2.0;
var SWPixel = new Coordinate(0, 0);
SWPixel.X = centerPixel.X - mapWidth / 2.0;
SWPixel.Y = centerPixel.Y + mapHeight / 2.0;
var NEWorld = new Coordinate(0, 0);
NEWorld.X = NEPixel.X / scale;
NEWorld.Y = NEPixel.Y / scale;
var SWWorld = new Coordinate(0, 0);
SWWorld.X = SWPixel.X / scale;
SWWorld.Y = SWPixel.Y / scale;
var NELatLon = InverseMercator(NEWorld.X, NEWorld.Y);
var SWLatLon = InverseMercator(SWWorld.X, SWWorld.Y);
return new MapCoordinates() { NorthEast = NELatLon, SouthWest = SWLatLon };
}
}
public class MapCoordinates
{
public Coordinate SouthWest { get; set; }
public Coordinate NorthEast { get; set; }
}
public class Coordinate
{
public double Latitude { get; set; }
public double Longitude { get; set; }
public double Y { get { return Latitude; } set { Latitude = value; } }
public double X { get { return Longitude; } set { Longitude = value; } }
public Coordinate(double lng, double lat)
{
Latitude = lat;
Longitude = lng;
}
public override string ToString()
{
return Math.Round(X, 6).ToString() + ", " + Math.Round(Y, 6).ToString();
}
}
输出：
SouthWest：51.438825,5.453483 NorthEast：51.452557,5.467153
SouthWest坐标与我自己的javascript翻译完全相同。因为peterjb在他的回答中声称他的解决方案适用于比利时的某个地区，所以我在相同的区域尝试了代码。然而，结果再次相当有点。
如此随机，我尝试使用尼日利亚一个名为Kano / Static map的城镇进行计算，令我惊讶的是，SouthWest角落非常准确！
输出：
SouthWest：12.015251,8.527824 NorthEast：12.028983,8.541404
之后我在世界上尝试了多个其他地区，但它往往很远，甚至有时甚至数百英里。所以也许尼日利亚的这个小镇只是巧合，但我希望有人能为我解释和/或解决方案。
答案 0 :(得分：0)
Coordinate类的构造函数接受（lng，lat）而不是（lat，lng），因为参数的顺序与传入的顺序不同。
我还注意到，Coordinate的ToString也会将打印覆盖为（lng，lat），因此如果您希望看到（lat，lng），可能会造成混淆。
只需在静态API中交换这两个。