Home >Backend Development >PHP Tutorial >How to query nearby people and their distance in PHP
the example in this article describes the implementation method of querying nearby people and their distance in php. share it with everyone for your reference, the details are as follows:
<?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 />';
}
?>