嗨，我需要计算经度和经度之间的距离。
我想避免对外部 API 的任何调用。
我试图在PHP中实现Haversine公式：
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;
}
}
用一些具有公共距离的给定点进行测试，我没有得到可靠的结果。
我不明白原始公式或我的实现中是否存在错误
不久前，我写了一个哈弗正弦公式的例子，并把它发布在我的网站上：
/**
* 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;
}
➽ 请注意，使用参数 $earthRadius 传入时，您可以以相同的单位获得距离。默认值为 6371000 米，因此结果也将以 [m] 为单位。例如，要以英里为单位获得结果，您可以以 $earthRadius 通过 3959 英里，结果将以 [mi] 为单位。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。
编辑：
正如TreyA正确指出的那样，由于舍入误差，Haversine公式在对跖点方面存在弱点(尽管它在小距离内是稳定的(。为了绕过它们，你可以改用文森蒂公式。
/**
* 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 * $earthRadius;
}
我找到了这段代码，它给了我可靠的结果。
function distance($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;
}
}
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>";
这只是
对@martinstoeckli的补充，@Janith钦塔纳回答。对于那些对哪种算法最快感到好奇的人，我编写了性能测试。最佳性能结果显示了 codexworld.com 优化的功能：
/**
* 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;
}
以下是测试结果：
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%
这里是计算两个纬度和经度之间距离的简单而完美的代码。从这里找到了以下代码 - http://www.codexworld.com/distance-between-two-addresses-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';
相当老的问题，但对于那些对返回与谷歌地图相同结果的PHP代码感兴趣的人，以下内容可以完成这项工作：
/**
* Computes the distance between two coordinates.
*
* Implementation based on reverse engineering of
* <code>google.maps.geometry.spherical.computeDistanceBetween()</code>.
*
* @param float $lat1 Latitude from the first point.
* @param float $lng1 Longitude from the first point.
* @param float $lat2 Latitude from the second point.
* @param float $lng2 Longitude from the second point.
* @param float $radius (optional) Radius in meters.
*
* @return float Distance in meters.
*/
function computeDistance($lat1, $lng1, $lat2, $lng2, $radius = 6378137)
{
static $x = M_PI / 180;
$lat1 *= $x; $lng1 *= $x;
$lat2 *= $x; $lng2 *= $x;
$distance = 2 * asin(sqrt(pow(sin(($lat1 - $lat2) / 2), 2) + cos($lat1) * cos($lat2) * pow(sin(($lng1 - $lng2) / 2), 2)));
return $distance * $radius;
}
我已经用各种坐标进行了测试，它工作得很好。
我认为它也应该比一些替代方案更快。但没有测试。
提示：谷歌地图使用6378137作为地球半径。因此，将其与其他算法一起使用也可能有效。
对于那些喜欢更短和更快的人(不调用 deg2rad(((。
function circle_distance($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
}
试试这个函数来计算到经纬度点之间的距离
function calculateDistanceBetweenTwoPoints($latitudeOne='', $longitudeOne='', $latitudeTwo='', $longitudeTwo='',$distanceUnit ='',$round=false,$decimalPoints='')
{
if (empty($decimalPoints))
{
$decimalPoints = '3';
}
if (empty($distanceUnit)) {
$distanceUnit = 'KM';
}
$distanceUnit = strtolower($distanceUnit);
$pointDifference = $longitudeOne - $longitudeTwo;
$toSin = (sin(deg2rad($latitudeOne)) * sin(deg2rad($latitudeTwo))) + (cos(deg2rad($latitudeOne)) * cos(deg2rad($latitudeTwo)) * cos(deg2rad($pointDifference)));
$toAcos = acos($toSin);
$toRad2Deg = rad2deg($toAcos);
$toMiles = $toRad2Deg * 60 * 1.1515;
$toKilometers = $toMiles * 1.609344;
$toNauticalMiles = $toMiles * 0.8684;
$toMeters = $toKilometers * 1000;
$toFeets = $toMiles * 5280;
$toYards = $toFeets / 3;
switch (strtoupper($distanceUnit))
{
case 'ML'://miles
$toMiles = ($round == true ? round($toMiles) : round($toMiles, $decimalPoints));
return $toMiles;
break;
case 'KM'://Kilometers
$toKilometers = ($round == true ? round($toKilometers) : round($toKilometers, $decimalPoints));
return $toKilometers;
break;
case 'MT'://Meters
$toMeters = ($round == true ? round($toMeters) : round($toMeters, $decimalPoints));
return $toMeters;
break;
case 'FT'://feets
$toFeets = ($round == true ? round($toFeets) : round($toFeets, $decimalPoints));
return $toFeets;
break;
case 'YD'://yards
$toYards = ($round == true ? round($toYards) : round($toYards, $decimalPoints));
return $toYards;
break;
case 'NM'://Nautical miles
$toNauticalMiles = ($round == true ? round($toNauticalMiles) : round($toNauticalMiles, $decimalPoints));
return $toNauticalMiles;
break;
}
}
然后将功能用作
echo calculateDistanceBetweenTwoPoints('11.657740','77.766270','11.074820','77.002160','ML',true,5);
希望对你有帮助
试试这个会得到很棒的结果
function getDistance($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);
}
对于确切的值，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;
}
嗯，我认为应该这样做...
编辑：
对于公式器和至少JS实现，请尝试： http://www.movable-type.co.uk/scripts/latlong.html
敢我...我忘了 deg2rad 圆函数中的所有值......
乘
数在每个坐标处都会改变，因为大圆距离理论写在这里：
http://en.wikipedia.org/wiki/Great-circle_distance
您可以使用此处描述的公式计算最接近的值：
http://en.wikipedia.org/wiki/Great-circle_distance#Worked_example
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'
你好在这里 使用两个不同的纬度和长距离获取距离和时间的代码
$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);
您可以查看下面的示例链接在php中使用纬度和经度获取两个不同位置之间的时间
最简单的方法之一是：
$my_latitude = "";
$my_longitude = "";
$her_latitude = "";
$her_longitude = "";
$distance = round((((acos(sin(($my_latitude*pi()/180)) * sin(($her_latitude*pi()/180))+cos(($my_latitude*pi()/180)) * cos(($her_latitude*pi()/180)) * cos((($my_longitude- $her_longitude)*pi()/180))))*180/pi())*60*1.1515*1.609344), 2);
echo $distance;
它将四舍五入到小数点后 2 位。
你可以试试库geo-math-php
composer require rkondratuk/geo-math-php:^1
例：
<?php
use PhpGeoMath'Model'Polar3dPoint;
$polarPoint1 = new Polar3dPoint(
40.758742779050706, -73.97855507715238, Polar3dPoint::EARTH_RADIUS_IN_METERS
);
$polarPoint2 = new Polar3dPoint(
40.74843388072615, -73.98566565776102, Polar3dPoint::EARTH_RADIUS_IN_METERS
);
$geoDistance = $polarPoint2->calcGeoDistanceToPoint($polarPoint1);