获取SW&;谷歌静态地图API的东北角
本文关键字:地图 API 东北角 静态 谷歌 SW amp 获取 | 更新日期: 2025-01-25 09:37:40
我在Unity应用程序中获取Google静态地图的边界时遇到问题。我已经尝试了在stackoverflow上找到的多个答案。
.NET相关问题
Javascript相关问题
我要计算的地图就是这个。
首先,我尝试将marcelo的代码从javascript翻译成C#。它被翻译成下面的类。
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);
}
}
所以一开始我试着只找到地图的西南角,距离很近,但有点远
MercatorProjection.GetCorners(new Coordinates(Coordinates.Longitude, Coordinates.Latitude), 16, 640, 640);
给我带来了这个结果。
由于有这么多人投票给了First,我试着把marcelo的答案翻译出来,我觉得我在把代码翻译成C#的过程中做错了什么。
然后我偶然发现了peterjb的答案,他已经将javascript答案翻译成了C#,甚至给出了他的代码达到谷歌静态地图正确边界的例子。因此,在下面的类中尝试了peterjb的代码之后;
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();
}
}
我再次尝试通过调用以下来计算西南角和东北角
Coordinates = new Coordinate(51.445691, 5.460318);
var result = GoogleMapsAPI.GetBounds(Coordinates, 16, 640, 640);
Debug.Log("SouthWest: "+result.SouthWest.ToString() + " NorthEast: "+result.NorthEast.ToString());
输出:
西南部:51.438825、5.453483东北部:51.452557、5.467153
西南坐标与我自己翻译的javascript完全相同。因此,由于peterjb在回答中声称他的解决方案适用于比利时的某个地区,我在同一地区尝试了该代码。然而,结果又有点不对劲
因此,我非常随机地尝试使用尼日利亚一个名为卡诺/静态地图的城镇的计算,令我惊讶的是,西南角非常准确!
输出:
西南部:12.015251、8.527824东北部:12.028983、8.541404
西南位置。
之后,我尝试了世界上其他多个地区,但通常都很远,有时甚至数百英里。所以,也许尼日利亚的这个小镇只是一个巧合,但我希望有人能给我一个解释和/或解决方案。
特别感谢peterjb和marcelo
Coordinate类的构造函数接受(lng,lat),而不是您所期望的(lat,lng),因此参数的顺序与您传入的顺序不同。
我还注意到Coordinate的ToString覆盖也打印为(lng,lat),所以如果你希望看到(lat,lng),这可能会令人困惑。
只需在静态API中交换这两个。