给定经纬度定位某个城市

时间：2015-01-12 14:24:41 阅读：281 评论：0 收藏：0 [点我收藏+]

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方案1

通过GEO HASH，抓取每个格子里的点的地址信息，然后把这个地址信息作为GEO中点坐标的一个信息，当查询给定经纬度是在哪个城市的时候，查询过程是查询距离这个经纬度最近的带有地址的点，认为给定经纬度和最近的这个点是同一个地址下的。通过不断完善GEO中的每个坐标下的地址信息，可以提高整个系统的精度。这种方法相当于一种间接获取的方法的。

方案2

我们知道每个地区是有一定区域的，可以认为是一个闭合的多边形，定位城市，可以理解为判断某个点，是否在某个区域内。

第一点：如何获取城市的区域边界

很幸运的是百度地图提供了这样的API

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>获取地区轮廓线</title>
<script type="text/javascript" src="http://api.map.baidu.com/api?v=1.3">
</script>
<style type="text/css">
body{font-size:13px;margin:10px}
#container{width:800px;height:500px;border:1px solid gray}
</style>
</head>
<body>
<div id="container"></div>
<br />
输入省、直辖市或县名称：<input type="text" id="districtName" style="width:80px" value="重庆市">
<input type="button" onclick="getBoundary()" value="获取轮廓线">

<script type="text/javascript">
var map = new BMap.Map("container");
map.centerAndZoom(new BMap.Point(116.403765, 39.914850), 5);
map.addControl(new BMap.NavigationControl({type: BMAP_NAVIGATION_CONTROL_SMALL}));
map.enableScrollWheelZoom();

function getBoundary(){       
    var bdary = new BMap.Boundary();
    var name = document.getElementById("districtName").value;
    bdary.get(name, function(rs){       //获取行政区域
        map.clearOverlays();        //清除地图覆盖物       
        var count = rs.boundaries.length; //行政区域的点有多少个
        for(var i = 0; i < count; i++){
            var ply = new BMap.Polygon(rs.boundaries[i], {strokeWeight: 2, strokeColor: "#ff0000"}); //建立多边形覆盖物
            map.addOverlay(ply);  //添加覆盖物
            map.setViewport(ply.getPath());    //调整视野         
        }                
    });   
}

</script>
</body>
</html>

View Code

如果没有这样的API，我们也可以变相获得，大多数地图提供了通过经纬度获取地址信息的API，我们可以取一个城市内的一点，然后在城市外周围再取N个点，我们通过二分的方法，可以不断逼近边界

PHP代码

function border($city, $p1, $p2) {
    $tpl = ‘http://api.map.baidu.com/geocoder/v2/?ak=你的应用key&callback=renderReverse&location=%s,%s&output=json&pois=1‘;
    $i = 5;
    while ($i>0) {
        $p1X = $p1[0];
        $p1Y = $p1[1];
        $p2X = $p2[0];
        $p2Y = $p2[1];
        $dX = abs($p1X - $p2X) / 2;
        $dY = abs($p1Y - $p2Y) / 2;
        $mX = min($p1X,$p2X) + $dX;
        $mY = min($p1Y,$p2Y) + $dY;

        $url = sprintf($tpl, $mY, $mX);
        $content = rtrim(strstr(file_get_contents($url), "{", false), ")");
        $data = json_decode($content, true);
        if (($data[‘result‘][‘addressComponent‘][‘city‘]) == $city) {
            $p1[0] = $mX;
            $p1[1] = $mY;
        } else {
            $p2[0] = $mX;
            $p2[1] = $mY;
        }
        $i--;
    }
    return array($p1[0],$p1[1]);
}

View Code

第二点：知道了区域，如何知道点在区域内

技术分享

左边的图是从某个点向待检测的点引射线，右边的图是从待检测的点，向某个点引射线，这两种方法都可以，判断射线与闭合区域每条边相交的次数，如果是奇数次的话，可以判断点在区域内，如果是偶数次的话，可以判断点不在区域内。

实际编程中，第二种方法实现起来会更简单一些，而且我们都是向X轴的正方向引射线会更简单。

第三点：如何检测某个点正方向的射线和某条边相交

技术分享

给定直线P1P2，判断点P平行于X轴正方的射线是否与直线P1P2相交？

首先，满足相交的条件，必须Py夹在P1y和P2y之间。

其次，满足相交的条件，Px小于P1P2最大x，不然肯定不想交了。

最后，P1,P2知道这两个点可以确定一条直线方程y=ax+b，然后p点正方的射线实际是y=Py的直线方程，这两条直线可以求出他们的焦点的x坐标xinters。如果Px小于xinters说明是在直线的y=ax+b的左边，向右引射线肯定能相交，如果Px大于xinters说明在y=ax+b的右边，向右引射线肯定不相交。

附上一段实现的代码

#define MIN(x,y) (x < y ? x : y)
#define MAX(x,y) (x > y ? x : y)
  
typedef struct {
   double x,y;
} Point;
  
int InsidePolygon(Point *polygon,int N,Point p)
{
  int counter = 0;
  int i;
  double xinters;
  Point p1,p2;
  
  p1 = polygon[0];
  for (i=1;i<=N;i++) {
    p2 = polygon[i % N];
    if (p.y > MIN(p1.y,p2.y)) { //低
      if (p.y <= MAX(p1.y,p2.y)) { //高
        if (p.x <= MAX(p1.x,p2.x)) { //右
          if (p1.y != p2.y) { //简单忽略平行X轴这种情况
            xinters = (p.y-p1.y)*(p2.x-p1.x)/(p2.y-p1.y)+p1.x; //交叉点坐标 参考./media/point-and-polygon/xinters.jpg
            if (p1.x == p2.x || p.x <= xinters)
              counter++;
          }
        }
      }
    }
    p1 = p2;
  }
  
  if (counter % 2 == 0)
    return 0;
  else
    return 1;
}

第四：幸运的是JAVA中的awt.Polygon已经实现了类似的算法

以下代码判断了一个经纬度是否在北京范围内

import java.awt.*;

public class Main {

    public static void init(Polygon p) {
        String s = "117.396814, 40.233095;117.360068, 40.242128;117.348757, 40.25027;117.343034, 40.283717;117.30242, 40.284012;117.300909, 40.298824;117.294971, 40.307248;117.280908, 40.31483;117.277872, 40.331475;117.281586, 40.338248;117.268763, 40.343838;117.248206, 40.376793;117.233105, 40.375358;117.230104, 40.381682;117.235062, 40.392322;117.241661, 40.395752;117.243127, 40.400548;117.246722, 40.40068;117.24604, 40.405045;117.238857, 40.410209;117.240581, 40.42306;117.256385, 40.435421;117.263364, 40.436067;117.26993, 40.447321;117.241196, 40.464298;117.242267, 40.475311;117.232341, 40.484311;117.234676, 40.48752;117.223574, 40.501212;117.215835, 40.502724;117.21653, 40.512643;117.220225, 40.518656;117.239555, 40.519506;117.245336, 40.523751;117.253813, 40.518445;117.266702, 40.525486;117.256848, 40.54366;117.259649, 40.547985;117.256124, 40.554638;117.272602, 40.564079;117.302711, 40.572713;117.317146, 40.582207;117.340905, 40.582418;117.356706, 40.586185;117.372601, 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39.731878;116.905895, 39.732316;116.918338, 39.737685;116.915713, 39.755851;116.905956, 39.765245;116.917067, 39.769397;116.926504, 39.785708;116.954922, 39.787312;116.958916, 39.788953;116.957802, 39.793369;116.947889, 39.802187;116.942966, 39.801267;116.941163, 39.807497;116.944047, 39.811272;116.934897, 39.8194;116.934864, 39.836027;116.921582, 39.8546;116.914208, 39.858458;116.910419, 39.856768;116.908704, 39.849417;116.914744, 39.836888;116.906097, 39.838156;116.892248, 39.85166;116.865868, 39.864936;116.858785, 39.876295;116.826594, 39.895721;116.810909, 39.894968;116.798184, 39.891052;116.790891, 39.904278;116.787847, 39.924116;116.790425, 39.947352;116.784383, 39.957745;116.785533, 39.962352;116.781118, 39.966493;116.778104, 39.962551;116.768888, 39.965436;116.766651, 39.963757;116.763082, 39.967772;116.774479, 39.991807;116.782163, 39.999264;116.777743, 40.020054;116.79169, 40.04063;116.808115, 40.035326;116.812623, 40.039475;116.829879, 40.034153;116.830104, 40.053132;116.856134, 40.058213;116.856655, 40.06071;116.862424, 40.059915;116.879125, 40.046781;116.887003, 40.052001;116.897445, 40.05203;116.91645, 40.058756;116.921103, 40.058581;116.924633, 40.052961;116.929155, 40.052608;116.937167, 40.060357;116.945078, 40.053844;116.949071, 40.054797;116.953609, 40.047448;116.968858, 40.057487;116.97638, 40.054388;116.979592, 40.047527;116.9842, 40.045447;117.01239, 40.038834;117.03162, 40.038702;117.045767, 40.055848;117.060658, 40.061359;117.062142, 40.064898;117.088476, 40.070829;117.095952, 40.076216;117.119763, 40.078627;117.142575, 40.070338;117.16233, 40.075069;117.167983, 40.081988;117.183281, 40.07842;117.191574, 40.091165;117.219451, 40.102784;117.230842, 40.100735;117.230798, 40.104818;117.256558, 40.125811;117.264635, 40.118738;117.282491, 40.119351;117.294419, 40.127966;117.305696, 40.128324;117.318401, 40.145199;117.351763, 40.148592;117.364739, 40.166702;117.360039, 40.179331;117.371849, 40.184583;117.39733, 40.184087;117.411835, 40.190451;117.413853, 40.193758;117.410394, 40.194931;117.402432, 40.192313;117.386786, 40.198355;117.387165, 40.205991;117.394824, 40.209093;117.395258, 40.212225;117.385567, 40.214483;117.38483, 40.223684;117.399491, 40.227136;117.396814, 40.233095";
        for ( String t : s.split(";")) {
            int x = (int) (Float.parseFloat(t.split(",")[0]) * 1000);
            int y = (int) (Float.parseFloat(t.split(",")[1]) * 1000);
            p.addPoint(x, y);
        }
    }

    public static boolean contains(Polygon p, float x, float y) {
        return p.contains(x * 1000, y * 1000);
    }

    public static void main(String[] args) {
        Polygon p = new Polygon();
        init(p);
        contains(p, 116.270313f, 39.909622f);
        contains(p, 115.590763f, 40.032598f);
        contains(p, 115.903517f ,40.875831f);
    }
}

给定经纬度定位某个城市

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原文地址：http://www.cnblogs.com/23lalala/p/4218156.html

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