PHP计算两个GPS点之间的距离

下面是PHP计算两个GPS点距离的函数：

PHP计算两个GPS点之间的距离
Mysql计算两GPS坐标的距离
javascript计算两个GPS点之间的距离

#lon为经度，lat为纬度，一定不要弄错了哦 function <span><a href="http://www.aiezu.com/tag/distance" title="查看 distance 中的全部文章" target="_blank" rel="nofollow">distance</a></span>($lon1, $lat1, $lon2, $lat2){ return (2*ATAN2(SQRT(SIN(($lat1-$lat2)*PI()/180/2) *SIN(($lat1-$lat2)*PI()/180/2)+ COS($lat2*PI()/180)*COS($lat1*PI()/180) *SIN(($lon1-$lon2)*PI()/180/2) *SIN(($lon1-$lon2)*PI()/180/2)), SQRT(1-SIN(($lat1-$lat2)*PI()/180/2) *SIN(($lat1-$lat2)*PI()/180/2) +COS($lat2*PI()/180)*COS($lat1*PI()/180) *SIN(($lon1-$lon2)*PI()/180/2) *SIN(($lon1-$lon2)*PI()/180/2))))*6378140; }

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#lon为经度，lat为纬度，一定不要弄错了哦 functiondistance($lon1,$lat1,$lon2,$lat2){ 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(($lon1-$lon2)*PI()/180/2) *SIN(($lon1-$lon2)*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(($lon1-$lon2)*PI()/180/2) *SIN(($lon1-$lon2)*PI()/180/2))))*6378140; }

调用方法：

echo distance(39.91917,116.3896,39.91726,116.3940);

1	echo distance(39.91917,116.3896,39.91726,116.3940);

以上就介绍了PHP计算两个GPS点之间的距离，包括了方面的内容，希望对PHP教程有兴趣的朋友有所帮助。