PHP查询附近的人及其距离的实现方法

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 - $dlng),
        'right-bottom'=>array('lat'=>$lat - $dlat, 'lng'=>$lng + $dlng)
        );
print_r($squares['left-top']['lat']);
//从数库查询匹配的记录
$info_sql = "select * from `A` where lat<>0 and lat>{$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;
  $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。