从Google静态地图获取像素协调 [英] Getting pixel coordinated from google static maps

问题描述

我想找出静态地图上经纬度的像素坐标.例如，我从以下位置下载了图像:

I want to find out the pixels coordinates of a lat/lng on an static map. For example i've downloaded an image from:

链接到图像

我想要的是纬度/经度，以便能够将纬度映射到像素坐标.我搜索了一下，发现墨卡托投影可以解决我的问题.但是我找不到任何合适的方法.有人能帮帮我吗.如URL所示，我也放大到了9.

What I want is from a lat/lng long to be able to map that latlng to pixel coordinates. I've searched a bit and found that mercator projection can solve my problem. However I could not find any proper way of doing it. Can somebody please help me. Also I've zoomed to 9 when as shown in the URL.

推荐答案

如果要直接在地图的位图上绘制，则将lat/Lon转换为google静态地图的像素非常有用.与通过URL传递数百个参数相比，这是一种更好的方法.我遇到了同样的问题，并在网上找到了四个解决方案，它们看起来非常相似，但是用其他语言编写.我将其翻译成C#.我敢肯定，在Java或C语言中也可以轻松使用此简单代码:

Converting Lat/Lon to pixel for a google static map is very useful if you want to draw on the map's bitmap directly. This can be a better approach than passing hundreds of parameters with the URL. I had the same problem and found four solutions in the web which looked very similar but where written in other languages. I translated it into C#. I am sure it will be easy to use this simple code also in Java or C:

//(half of the earth circumference's in pixels at zoom level 21)
static double offset = 268435456; 
static double radius = offset / Math.PI;
// X,Y ... location in degrees
// xcenter,ycenter ... center of the map in degrees (same value as in 
// the google static maps URL)
// zoomlevel (same value as in the google static maps URL)
// xr, yr and the returned Point ... position of X,Y in pixels relativ 
// to the center of the bitmap
static Point Adjust(double X, double Y, double xcenter, double ycenter, 
                    int zoomlevel)
{
    int xr = (LToX(X) - LToX(xcenter)) >> (21 - zoomlevel);
    int yr = (LToY(Y) - LToY(ycenter)) >> (21 - zoomlevel);
    Point p = new Point(xr, yr);
    return p;
}

static int LToX(double x)
{
    return (int)(Math.Round(offset + radius * x * Math.PI / 180));
}

static int LToY(double y)
{
    return (int)(Math.Round(offset - radius * Math.Log((1 + 
                 Math.Sin(y * Math.PI / 180)) / (1 - Math.Sin(y * 
                 Math.PI / 180))) / 2));
}

用法:

1. 调用此函数以获取X和Y像素坐标
2. 结果参考到位图的中心，因此添加 位图将width/2和heigth/2映射到x和y值.这给你 绝对像素位置
3. 检查像素位置是否在您的位图之内
4. 随心所欲画

1. call this function to get the X and Y pixel coordinates
2. the result is referenced to the center of the bitmap, so add the bitmaps width/2 and heigth/2 to the x and y value. This gives you the absolute pixel position
3. check if the pixel position is inside your bitmap
4. draw whatever you want

由于Google墨卡托投影的变体，它无法在极点附近工作，但是对于通常的坐标来说，效果很好.

It does not work close to the poles because of Google's variant of Mercator projection, but for the usual coordinates it works very well.

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