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:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
|
<?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。