我想在php中使用这个公式。我有一个保存了纬度和经度值的数据库。
我想找出,在输入一定的纬度和经度值的情况下,从数据库中的每个点到这个点的所有距离(以公里为单位)。为此,我在googlemaps api上使用了以下公式:
( 6371 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) )
当然,在php中,我将弧度替换为deg2rad
输入值37,-122是我的.The值,而lng是我在数据库中的值。
下面是我的代码。问题是出了什么问题,但我不明白是什么。距离的值当然是错误的。
//values of latitude and longitute in input (Rome - eur, IT)
$center_lat = "41.8350";
$center_lng = "12.470";
//connection to database. it works
(..)
//to take each value in the database:
$query = "SELECT * FROM Dati";
$result = mysql_query($query);
while ($row = @mysql_fetch_assoc($result)){
$lat=$row['Lat']);
$lng=$row['Lng']);
$distance =( 6371 * acos((cos(deg2rad($center_lat)) ) * (cos(deg2rad($lat))) * (cos(deg2rad($lng) - deg2rad($center_lng)) )+ ((sin(deg2rad($center_lat))) * (sin(deg2rad($lat))))) );
}
对于值,例如:$lat= 41.9133741000 $lng= 12.5203944000
我有distance="4826.9341106926“的输出
发布于 2015-06-08 15:09:31
public function getDistanceBetweenTwoPoints($point1 , $point2){
// array of lat-long i.e $point1 = [lat,long]
$earthRadius = 6371; // earth radius in km
$point1Lat = $point1[0];
$point2Lat =$point2[0];
$deltaLat = deg2rad($point2Lat - $point1Lat);
$point1Long =$point1[1];
$point2Long =$point2[1];
$deltaLong = deg2rad($point2Long - $point1Long);
$a = sin($deltaLat/2) * sin($deltaLat/2) + cos(deg2rad($point1Lat)) * cos(deg2rad($point2Lat)) * sin($deltaLong/2) * sin($deltaLong/2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$distance = $earthRadius * $c;
return $distance; // in km
}
发布于 2013-02-07 19:48:53
function getDistance($latitude1, $longitude1, $latitude2, $longitude2) {
$earth_radius = 6371;
$dLat = deg2rad($latitude2 - $latitude1);
$dLon = deg2rad($longitude2 - $longitude1);
$a = sin($dLat/2) * sin($dLat/2) + cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * sin($dLon/2) * sin($dLon/2);
$c = 2 * asin(sqrt($a));
$d = $earth_radius * $c;
return $d;
}
正如您所看到的,这与您的代码有许多不同之处。我不知道你是否有不同的方法来处理这个公式,或者当转换到PHP时可能有一些步骤出错,但是上面的公式应该是有效的。
发布于 2013-02-07 20:06:33
我使用以下存储过程直接计算查询内部的距离:
CREATE FUNCTION GEODIST (lat1 DOUBLE, lon1 DOUBLE, lat2 DOUBLE, lon2 DOUBLE)
RETURNS DOUBLE
DETERMINISTIC
BEGIN
DECLARE dist DOUBLE;
SET dist = round(acos(cos(radians(lat1))*cos(radians(lon1))*cos(radians(lat2))*cos(radians(lon2)) + cos(radians(lat1))*sin(radians(lon1))*cos(radians(lat2))*sin(radians(lon2)) + sin(radians(lat1))*sin(radians(lat2))) * 6378.8, 1);
RETURN dist;
END|
您只需在phpMyAdmin中将上述代码作为SQl语句执行,即可创建该过程。只需注意结尾的|,因此在SQL输入窗口中,选择|符号作为limiter。
然后在查询中,像这样调用它:
$sql = "
SELECT `locations`.`name`, GEODIST(`locations`.`lat`, `locations`.`lon`, " . $lat_to_calculate . ", " . $lon_to_calculate . ") AS `distance`
FROM `locations` ";
我发现这比在运行查询后在PHP中计算它要快得多。
https://stackoverflow.com/questions/14750275
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