测量PHP中两个坐标之间的距离。

Hi I have the need to calculate the distance between two points having the lat and long.

你好，我需要计算两点之间的距离。

I would like to avoid any call to external API.

我希望避免调用任何外部API。

I tried to implement the Haversine Formula in PHP:

我尝试在PHP中实现Haversine公式:

Here is the code:

这是代码:

class CoordDistance
 {
    public $lat_a = 0;
    public $lon_a = 0;
    public $lat_b = 0;
    public $lon_b = 0;

    public $measure_unit = 'kilometers';

    public $measure_state = false;

    public $measure = 0;

    public $error = '';



    public function DistAB()

      {
          $delta_lat = $this->lat_b - $this->lat_a ;
          $delta_lon = $this->lon_b - $this->lon_a ;

          $earth_radius = 6372.795477598;

          $alpha    = $delta_lat/2;
          $beta     = $delta_lon/2;
          $a        = sin(deg2rad($alpha)) * sin(deg2rad($alpha)) + cos(deg2rad($this->lat_a)) * cos(deg2rad($this->lat_b)) * sin(deg2rad($beta)) * sin(deg2rad($beta)) ;
          $c        = asin(min(1, sqrt($a)));
          $distance = 2*$earth_radius * $c;
          $distance = round($distance, 4);

          $this->measure = $distance;

      }
    }

Testing it with some given points which have public distances I don't get a reliable result.

用一些有公共距离的给定的点来测试它，我没有得到一个可靠的结果。

I don't understand if there is an error in the original formula or in my implementation

我不明白原始公式或我的实现中是否有错误

9 个解决方案

#1

Not long ago I wrote an example of the haversine formula, and published it on my website:

不久前，我写了一个haversine公式的例子，并在我的网站上发表:

/**
 * 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;
}

➽ Note that you get the distance back in the same unit as you pass in with the parameter $earthRadius. The default value is 6371000 meters so the result will be in [m] too. To get the result in miles, you could e.g. pass 3959 miles as $earthRadius and the result would be in [mi]. In my opinion it is a good habit to stick with the SI units, if there is no particular reason to do otherwise.

➽注意得到的距离在同一单位作为你传入参数earthRadius美元。默认值是6371000米，所以结果也在[m]中。为了得到英里数的结果，你可以通过3959英里作为$earthRadius，结果将在[mi]。在我看来，如果没有特别的理由，坚持使用SI单位是一个好习惯。

Edit:

编辑:

As TreyA correctly pointed out, the Haversine formula has weaknesses with antipodal points because of rounding errors (though it is stable for small distances). To get around them, you could use the Vincenty formula instead.

正如TreyA所指出的那样，由于舍入误差(虽然对于小距离来说是稳定的)，Haversine公式在反对点上存在弱点。为了避开它们，你可以使用Vincenty公式。

/**
 * Calculates the great-circle distance between two points, with
 * the Vincenty 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)
 */
public static function vincentyGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $lonDelta = $lonTo - $lonFrom;
  $a = pow(cos($latTo) * sin($lonDelta), 2) +
    pow(cos($latFrom) * sin($latTo) - sin($latFrom) * cos($latTo) * cos($lonDelta), 2);
  $b = sin($latFrom) * sin($latTo) + cos($latFrom) * cos($latTo) * cos($lonDelta);

  $angle = atan2(sqrt($a), $b);
  return $angle * $earthRadius;
}

#2

I found this code which is giving me reliable results.

我找到了这个代码，它给了我可靠的结果。

function distance($lat1, $lon1, $lat2, $lon2, $unit) {

  $theta = $lon1 - $lon2;
  $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
  $dist = acos($dist);
  $dist = rad2deg($dist);
  $miles = $dist * 60 * 1.1515;
  $unit = strtoupper($unit);

  if ($unit == "K") {
      return ($miles * 1.609344);
  } else if ($unit == "N") {
      return ($miles * 0.8684);
  } else {
      return $miles;
  }
}

results :

结果:

echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br>";

#3

It's just addition to @martinstoeckli and @Janith Chinthana answers. For those who curious about which algorithm is fastest i wrote the performance test. Best performance result shows optimized function from codexworld.com:

@martinstoeckli和@Janith Chinthana回答。对于那些想知道哪种算法最快的人，我写了性能测试。最佳性能结果显示codexworld.com优化功能:

/**
 * Optimized algorithm from http://www.codexworld.com
 *
 * @param float $latitudeFrom
 * @param float $longitudeFrom
 * @param float $latitudeTo
 * @param float $longitudeTo
 *
 * @return float [km]
 */
function codexworldGetDistanceOpt($latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo)
{
    $rad = M_PI / 180;
    //Calculate distance from latitude and longitude
    $theta = $longitudeFrom - $longitudeTo;
    $dist = sin($latitudeFrom * $rad) * sin($latitudeTo * $rad) +  cos($latitudeFrom * $rad) * cos($latitudeTo * $rad) * cos($theta * $rad);

    return acos($dist) / $rad * 60 *  1.853;
}

Here is test results:

这是测试结果:

Test name       Repeats         Result          Performance     
codexworld-opt  10000           0.084952 sec    +0.00%
codexworld      10000           0.104127 sec    -22.57%
custom          10000           0.107419 sec    -26.45%
custom2         10000           0.111576 sec    -31.34%
custom1         10000           0.136691 sec    -60.90%
vincenty        10000           0.165881 sec    -95.26%

#4

Here the simple and perfect code for calculating the distance between two latitude and longitude. The following code have been found from here - http://www.codexworld.com/distance-between-two-addresses-google-maps-api-php/

这里是计算纬度和经度之间距离的简单而完美的代码。以下代码已经在这里找到了:http://www.codexworld.com/distance- two- addresss-google -maps-api-php/。

$latitudeFrom = '22.574864';
$longitudeFrom = '88.437915';

$latitudeTo = '22.568662';
$longitudeTo = '88.431918';

//Calculate distance from latitude and longitude
$theta = $longitudeFrom - $longitudeTo;
$dist = sin(deg2rad($latitudeFrom)) * sin(deg2rad($latitudeTo)) +  cos(deg2rad($latitudeFrom)) * cos(deg2rad($latitudeTo)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;

$distance = ($miles * 1.609344).' km';

#5

For the ones who like shorter and faster(not calling deg2rad()).

对于喜欢更短、更快(不调用deg2rad())的人。

function circle_distance($lat1, $lon1, $lat2, $lon2) {
  $rad = M_PI / 180;
  return acos(sin($lat2*$rad) * sin($lat1*$rad) + cos($lat2*$rad) * cos($lat1*$rad) * cos($lon2*$rad - $lon1*$rad)) * 6371;// Kilometers
}

#6

For exact values do it like that:

对于精确的值，这样做:

public function DistAB()
{
      $delta_lat = $this->lat_b - $this->lat_a ;
      $delta_lon = $this->lon_b - $this->lon_a ;

      $a = pow(sin($delta_lat/2), 2);
      $a += cos(deg2rad($this->lat_a9)) * cos(deg2rad($this->lat_b9)) * pow(sin(deg2rad($delta_lon/29)), 2);
      $c = 2 * atan2(sqrt($a), sqrt(1-$a));

      $distance = 2 * $earth_radius * $c;
      $distance = round($distance, 4);

      $this->measure = $distance;
}

Hmm I think that should do it...

嗯，我觉得应该这样……

Edit:

编辑:

For formulars and at least JS-implementations try: http://www.movable-type.co.uk/scripts/latlong.html

对于公式和至少jsp实现，请尝试:http://www.movable-type.co.uk/scripts/latlong.html

Dare me... I forgot to deg2rad all the values in the circle-functions...

敢我……我忘了deg2rad所有的循环函数值…

#7

Hello here Code For Get Distance and Time Using Two Different Lat and Long

你好，这里的代码获取距离和时间使用两个不同的Lat和Long

$url ="https://maps.googleapis.com/maps/api/distancematrix/json?units=imperial&origins=16.538048,80.613266&destinations=23.0225,72.5714";



    $ch = curl_init();
    // Disable SSL verification

    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
    // Will return the response, if false it print the response
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    // Set the url
    curl_setopt($ch, CURLOPT_URL,$url);
    // Execute
    $result=curl_exec($ch);
    // Closing
    curl_close($ch);

    $result_array=json_decode($result);
print_r($result_array);

You can check Example Below Link get time between two different locations using latitude and longitude in php

在php中，可以使用纬度和经度检查链接下面的示例获取两个不同位置之间的时间

#8

Try this gives awesome results

试试这个，会有很棒的结果

function getDistance($point1_lat, $point1_long, $point2_lat, $point2_long, $unit = 'km', $decimals = 2) {
        // Calculate the distance in degrees
        $degrees = rad2deg(acos((sin(deg2rad($point1_lat))*sin(deg2rad($point2_lat))) + (cos(deg2rad($point1_lat))*cos(deg2rad($point2_lat))*cos(deg2rad($point1_long-$point2_long)))));

        // Convert the distance in degrees to the chosen unit (kilometres, miles or nautical miles)
        switch($unit) {
            case 'km':
                $distance = $degrees * 111.13384; // 1 degree = 111.13384 km, based on the average diameter of the Earth (12,735 km)
                break;
            case 'mi':
                $distance = $degrees * 69.05482; // 1 degree = 69.05482 miles, based on the average diameter of the Earth (7,913.1 miles)
                break;
            case 'nmi':
                $distance =  $degrees * 59.97662; // 1 degree = 59.97662 nautic miles, based on the average diameter of the Earth (6,876.3 nautical miles)
        }
        return round($distance, $decimals);
    }

#9

The multiplier is changed at every coordinate because of the great circle distance theory as written here :

由于大圆距离理论，乘法器在每个坐标上都发生了变化。

http://en.wikipedia.org/wiki/Great-circle_distance

http://en.wikipedia.org/wiki/Great-circle_distance

and you can calculate the nearest value using this formula described here:

你可以用这个公式计算最接近的值

http://en.wikipedia.org/wiki/Great-circle_distance#Worked_example

http://en.wikipedia.org/wiki/Great-circle_distance Worked_example

the key is converting each degree - minute - second value to all degree value:

关键是将每个度-分-秒值转换为所有度值:

N 36°7.2', W 86°40.2'  N = (+) , W = (-), S = (-), E = (+) 
referencing the Greenwich meridian and Equator parallel

(phi)     36.12° = 36° + 7.2'/60' 

(lambda)  -86.67° = 86° + 40.2'/60'