PHP查询附近人及其距离实现方法

PHP查询附近人及其距离实现方法 ?精品文档? PHP查询附近人及其距离实现方法 本文实例讲述了PHP查询附近的人及其距离的实现方法。分享给大家供大家参考,具体如下, <?php //获取该点周围的4个点 $distance = 1;//范围,单位千米, $lat = 113.873643; $lng = 22.573969; define(‘EARTH_RADIUS’, 6371);//地球半径,平均半径为6371km $dlng = 2 * asin(sin($distance / (2 * EARTH_RADIUS)) / cos(deg2rad($lat))); $dlng = rad2deg($dlng); $dlat = $distance/EARTH_RADIUS; $dlat = rad2deg($dlat); $squares = array(‘left-top’=array(‘lat’=$lat + $dlat,’lng’=$lng-$dlng),         ‘right-top’=array(‘lat’=$lat + $dlat, ‘lng’=$lng + $dlng),         ‘left-bottom’=array(‘lat’=$lat - $dlat, ‘lng’=$lng - $dlng),         1 / 3 ?精品文档? ‘right-bottom’=array(‘lat’=$lat - $dlat, ‘lng’=$lng + $dlng)         ); print_r($squares[‘left-top’][‘lat’]); //从数库查询匹配的 $info_sql = “select * from `A` where lat<0 and lat{$squares[‘right-bottom’][‘lat’]} and lat<{$squares[‘left-top’][‘lat’]} and lng{$squares[‘left-top’][‘lng’]} and lng<{$squares[‘right-bottom’][‘lng’]} “; //获取两点之间的距离 function getDistanceBetweenPointsNew($latitude1, $longitude1, $latitude2, $longitude2) {   $theta = $longitude1 - $longitude2;   $miles = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) + (cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * cos(deg2rad($theta)));   $miles = acos($miles);   $miles = rad2deg($miles);   $miles = $miles * 60 * 1.1515;   $feet = $miles * 5280;   $yards = $feet / 3;   $kilometers = $miles * 1.609344; 2 / 3 ?精品文档?   $meters = $kilometers * 1000;   return compact(‘miles’,’feet’,’yards’,’kilometers’,’meters’);  } $point1 = array(‘lat’ = 40.770623, ‘long’ = -73.964367); $point2 = array(‘lat’ = 40.758224, ‘long’ = -73.917404); $distance = getDistanceBetweenPointsNew($point1[‘lat’], $point1[‘long’], $point2[‘lat’], $point2[‘long’]); foreach ($distance as $unit = $value) {   echo $unit.’: ‘.number_format($value,4).’<br /’; } ? 希望本文所述对大家PHP程序设计有所帮助。 3 / 3