24

我想在 php 中使用这个公式。我有一个保存了一些纬度和经度值的数据库。

我想通过输入中的某个纬度和经度值来查找从该点到数据库中每个点的所有距离(以公里为单位)。为此,我使用了 googlemaps api 上的公式:

( 6371 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) )

当然在 php 中使用它我用deg2rad.The 值替换了弧度。值 37,-122 是我的输入值,而 lat,lng 是我在数据库中的值。

下面是我的代码。问题是出了点问题,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))))) );
    }

例如值: $lat= 41.9133741000 $lng= 12.5203​​944000

我有 distance="4826.9341106926" 的输出

4

5 回答 5

59

您使用的公式似乎是反余弦而不是半正弦公式。hasrsine 公式确实更适合计算球体上的距离,因为它不太容易出现舍入误差。

/**
 * Calculates the great-circle distance between two points, with
 * the Haversine formula.
 * @param float $latitudeFrom Latitude of start point in [deg decimal]
 * @param float $longitudeFrom Longitude of start point in [deg decimal]
 * @param float $latitudeTo Latitude of target point in [deg decimal]
 * @param float $longitudeTo Longitude of target point in [deg decimal]
 * @param float $earthRadius Mean earth radius in [m]
 * @return float Distance between points in [m] (same as earthRadius)
 */
function haversineGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $latDelta = $latTo - $latFrom;
  $lonDelta = $lonTo - $lonFrom;

  $angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
    cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
  return $angle * $earthRadius;
}

PS我在你的代码中找不到错误,所以这只是你写的错字$lat= 41.9133741000 $lat= 12.5203944000吗?也许你只是用 $lat=12.5203​​944000 和 $long=0 计算,因为你覆盖了你的 $lat 变量。

编辑:

测试了代码,它返回了正确的结果:

$center_lat = 41.8350;
$center_lng = 12.470;
$lat = 41.9133741000;
$lng = 12.5203944000;

// test with your arccosine formula
$distance =( 6371 * acos((cos(deg2rad($center_lat)) ) * (cos(deg2rad($lat))) * (cos(deg2rad($lng) - deg2rad($center_lng)) )+ ((sin(deg2rad($center_lat))) * (sin(deg2rad($lat))))) );
print($distance); // prints 9.662174538188

// test with my haversine formula
$distance = haversineGreatCircleDistance($center_lat, $center_lng, $lat, $lng, 6371);
print($distance); // prints 9.6621745381693
于 2013-02-07T13:04:11.507 回答
6
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
}
于 2015-06-08T07:09:31.187 回答
4

这个链接

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,这与您的代码有很多不同。我不知道您是否对公式有不同的方法,或者转换为 PHP 时的某个步骤出错了,但是上面的公式应该可以工作。

于 2013-02-07T11:48:52.870 回答
1

我使用以下存储过程直接在查询中计算距离:

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|

您只需从 phpMyAdmin 中将上述内容作为 SQl 语句执行以创建过程。请注意结尾 |,因此在您的 SQL 输入窗口中,选择 | 作为限制器签名。

然后在查询中,这样调用它:

$sql = "
SELECT `locations`.`name`, GEODIST(`locations`.`lat`, `locations`.`lon`, " . $lat_to_calculate . ", " . $lon_to_calculate . ") AS `distance`
FROM `locations` ";

我发现这比运行查询后在 PHP 中计算要快得多。

于 2013-02-07T12:06:33.433 回答
1

我制作了具有四个参数的静态函数 getDistance 的 hasrsine 类,它返回与机器人位置点的距离

class HaverSign {
    
     public static 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;
}
}

上面的类存放在根目录,根目录包含classes文件夹在任何php页面中使用以下方式调用

include "../classes/HaverSign.php";
$haversign=new HaverSign();

$lat=18.5204;
$lon=73.8567;

$lat1=18.5404;
$lon1=73.8167;

$dist = $haversign->getDistance($lat,$lon,$lat1,$lon1);
echo $dist;

输出如下

4.7676529976827
于 2016-04-01T06:03:33.947 回答