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:
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<?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 />' ;
}
?>
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希望本文所述对大家PHP程序设计有所帮助。