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。

<?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 />';
}
?>

链接：https://www.php.cn/php-weizijiaocheng-316.html