java 经纬度两点 距离或范围
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 34.264648,108.952736,39.904549,116.407288 double 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.
oc中
//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