计算两个经纬度之间的距离
PHP计算：
#lng为经度，lat为纬度，一定不要弄错了哦
function distance($lat1, $lng1, $lat2, $lng2){
return (2*ATAN2(SQRT(SIN(($lat1-$lat2)*PI()/180/2)
*SIN(($lat1-$lat2)*PI()/180/2)+
COS($lat2*PI()/180)*COS($lat1*PI()/180)
*SIN(($lng1-$lng2)*PI()/180/2)
*SIN(($lng1-$lng2)*PI()/180/2)),
SQRT(1-SIN(($lat1-$lat2)*PI()/180/2)
*SIN(($lat1-$lat2)*PI()/180/2)
+COS($lat2*PI()/180)*COS($lat1*PI()/180)
*SIN(($lng1-$lng2)*PI()/180/2)
*SIN(($lng1-$lng2)*PI()/180/2))))*6378140;
}mysql计算：
set @lng1=116.3899; set @lat1=39.91578; set @lng2=116.3904; set @lat2=39.91576; select (2*ATAN2(SQRT(SIN((@lat1-@lat2)*PI()/180/2) *SIN((@lat1-@lat2)*PI()/180/2)+ COS(@lat2*PI()/180)*COS(@lat1*PI()/180) *SIN((@lng1-@lng2)*PI()/180/2) *SIN((@lng1-@lng2)*PI()/180/2)), SQRT(1-SIN((@lat1-@lat2)*PI()/180/2) *SIN((@lat1-@lat2)*PI()/180/2) +COS(@lat2*PI()/180)*COS(@lat1*PI()/180) *SIN((@lng1-@lng2)*PI()/180/2) *SIN((@lng1-@lng2)*PI()/180/2))))*6378140;
google给的解决方案（基于公里km）
SELECT geo_id, `name`,( 6371 * acos ( cos ( radians(33.958887) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(118.302416) ) + sin ( radians(33.958887) ) * sin( radians( lat ) ) ) ) AS distance FROM geo HAVING distance < 20 ORDER BY distance LIMIT 0 , 20;
javascript方案：
function getDisance(lat1, lng1, lat2, lng2) { // lat为纬度, lng为经度, 一定不要弄错
function toRad(d) { return d * Math.PI / 180; }
var dis = 0;
var radLat1 = toRad(lat1);
var radLat2 = toRad(lat2);
var deltaLat = radLat1 - radLat2;
var deltaLng = toRad(lng1) - toRad(lng2);
var dis = 2 * Math.asin(Math.sqrt(Math.pow(Math.sin(deltaLat / 2), 2) + Math.cos(radLat1) * Math.cos(radLat2) * Math.pow(Math.sin(deltaLng / 2), 2)));
return dis * 6378137;
}
alert( getDisance(39.91917,116.3896,39.91726,116.3940) );
