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中纬度两点距离|范围

2012-06-28 
经纬度两点距离|范围!愁人!第一弄这个.网上找到不少算法,但实用的没几个,而且说的有些抽象,整理了一下分享

经纬度两点距离|范围!
愁人!第一弄这个.网上找到不少算法,但实用的没几个,而且说的有些抽象,整理了一下分享!

public class LatLonUtil {    private static final double PI = 3.14159265;    //老祖真理    private static final double EARTH_RADIUS = 6378137;    //赤道半径    private static final double RAD = Math.PI / 180.0;    //@see http://snipperize.todayclose.com/snippet/php/SQL-Query-to-Find-All-Retailers-Within-a-Given-Radius-of-a-Latitude-and-Longitude--65095/ //The circumference of the earth is 24,901 miles.    //24,901/360 = 69.17 miles / degree      /**     * @param raidus 单位米     * return minLat,minLng,maxLat,maxLng     */    public static double[] getAround(double lat,double lon,int raidus){Double latitude = lat;Double longitude = lon;Double degree = (24901*1609)/360.0;double raidusMile = raidus;Double dpmLat = 1/degree;Double radiusLat = dpmLat*raidusMile;Double minLat = latitude - radiusLat;Double maxLat = latitude + radiusLat;Double mpdLng = degree*Math.cos(latitude * (PI/180));Double dpmLng = 1 / mpdLng;Double radiusLng = dpmLng*raidusMile;Double minLng = longitude - radiusLng;Double maxLng = longitude + radiusLng;//System.out.println("["+minLat+","+minLng+","+maxLat+","+maxLng+"]");return new double[]{minLat,minLng,maxLat,maxLng};}        /**     * 根据两点间经纬度坐标(double值),计算两点间距离,单位为米     * @param lng1     * @param lat1     * @param lng2     * @param lat2     * @return     */    public static double getDistance(double lng1, double lat1, double lng2, double lat2)    {       double radLat1 = lat1*RAD;       double radLat2 = lat2*RAD;       double a = radLat1 - radLat2;       double b = (lng1 - lng2)*RAD;       double s = 2 * Math.asin(Math.sqrt(Math.pow(Math.sin(a/2),2) +        Math.cos(radLat1)*Math.cos(radLat2)*Math.pow(Math.sin(b/2),2)));       s = s * EARTH_RADIUS;       s = Math.round(s * 10000) / 10000;       return s;    }public static void main(String[] args){Double lat1 = 34.264648;Double lon1 = 108.952736;int radius = 1000;//[34.25566276027792,108.94186385411045,34.27363323972208,108.96360814588955]getAround(lat1,lon1,radius);//911717.0   34.264648,108.952736,39.904549,116.407288double dis = getDistance(108.952736,34.264648,116.407288,39.904549); System.out.println(dis);}}

以上,最大经度,纬度,最小经度,纬度! 两点之间的距离
以上引用别人,写的挺不错,但是有些麻烦!过去结果不够直接.
    //object-c,获取搜索目标的经纬度        var geocoder = new GClientGeocoder();          geocoder.getLocations(address, function($){          var lalton = $.Placemark[0].Point.coordinates;          areaMap(lalton[1],lalton[0]);          //alert(lalton[1] + "," + lalton[0]);      });  

注意这里数据库是mysql
    SELECT id, ( 3959 * acos( cos( radians(lat_t) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(lng_t) )      + sin( radians(lat_t) ) * sin( radians( lat ) ) ) ) AS distance      FROM Stores HAVING distance < 25      ORDER BY distance      LIMIT 0,20  

SELECT * from app_activity where (3959*acos(cos(radians(36.0971114))*cos(radians(latitude))*cos(radians(longitude)-radians(103.6130684))+sin(radians(36.0971114))*sin(radians(latitude)))) <1

1、3959为地球半径
2、25为搜索半径
3、3959和25都是以“英里”为单位如果需要改成“公里”来计算的,都需要乘以1.60931

select * from location where sqrt(          (           ((113.914619-longitude)*PI()*12656*cos(((22.50128+latitude)/2)*PI()/180)/180)           *           ((113.914619-longitude)*PI()*12656*cos (((22.50128+latitude)/2)*PI()/180)/180)          )          +          (           ((22.50128-latitude)*PI()*12656/180)           *           ((22.50128-latitude)*PI()*12656/180)          ) 

经度:113.914619
纬度:22.50128
范围:2km
longitude为数据表经度字段
latitude为数据表纬度字段


地球在线http://www.earthol.com/获取经纬度做测试

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