PHP查询附近的人及其距离的实现方法
2016-08-23来源:

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。

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