php 在PHP中测量两个坐标之间的距离
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Measuring the distance between two coordinates in PHP
提问by maxdangelo
Hi I have the need to calculate the distance between two points having the lat and long.
嗨，我需要计算具有纬度和经度的两点之间的距离。
I would like to avoid any call to external API.
我想避免对外部 API 的任何调用。
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点测试它我没有得到可靠的结果。
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
回答by martinstoeckli
Not long ago I wrote an example of the haversine formula, and published it on my website:
不久前我写了一个半正弦公式的例子，并发布在我的网站上：
/**
* 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;
}
? 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 $earthRadiusand the result would be in [mi]. In my opinion it is a good habit to stick with the SI units, if there is no particular reason to do otherwise.
? 请注意，返回的距离与使用参数传入的单位相同$earthRadius。默认值为 6371000 米，因此结果也会以 [m] 为单位。要获得以英里为单位的结果，您可以例如通过 3959 英里，$earthRadius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。
Edit:
编辑：
As TreyA correctly pointed out, the Haversine formula has weaknesses with antipodal pointsbecause of rounding errors (though it isstable for small distances). To get around them, you could use the Vincenty formulainstead.
正如 TreyA 正确指出的那样，由于舍入误差，Haversine 公式在对映点上存在弱点（尽管它对于小距离是稳定的）。要绕过它们，您可以改用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 * $earthRadius;
}
回答by Janith Chinthana
I found this codewhich is giving me reliable results.
我发现这段代码给了我可靠的结果。
function distance($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>";
回答by Alexander Yancharuk
It's just addition to @martinstoeckliand @Janith Chinthanaanswers. For those who curious about which algorithm is fastest i wrote the performance test. Best performance result shows optimized function from codexworld.com:
这只是@martinstoeckli和@Janith Chinthana答案的补充。对于那些对哪种算法最快的人感到好奇，我编写了性能测试。最佳性能结果显示来自codexworld.com 的优化功能：
/**
* 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:
下面是测试结果：
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%
回答by JoyGuru
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/
这里是计算两个经纬度之间距离的简单而完美的代码。从这里找到以下代码 - 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';
回答by Semra
For the ones who like shorter and faster(not calling deg2rad()).
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
}
回答by Diego Andrade
Quite old question, but for those interested in a PHP code that returns the same results as Google Maps, the following does the job:
很老的问题，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 * $radius;
}
I've tested with various coordinates and it works perfectly.
我已经用各种坐标进行了测试，效果很好。
I think it should be faster then some alternatives too. But didn't tested that.
我认为它也应该比一些替代品更快。但没有测试过。
Hint: Google Maps uses 6378137 as Earth radius. So using it with other algorithms might work as well.
提示：谷歌地图使用 6378137 作为地球半径。因此，将它与其他算法一起使用也可能有效。
回答by Amit
Try this gives awesome results
试试这个会产生很棒的结果
function getDistance($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);
}
回答by Jydipsinh Parmar
Hello here Code For Get Distance and Time Using Two Different Lat and Long
你好这里使用两个不同的纬度和经度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($result_array);
You can check Example Below Link get time between two different locations using latitude and longitude in php
您可以使用php中的纬度和经度检查下面的示例链接获取两个不同位置之间的时间
回答by Manojkiran.A
Try this function out to calculate distance between to points of latitude and longitude
试试这个函数来计算纬度和经度点之间的距离
function calculateDistanceBetweenTwoPoints($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;
}
}
Then use the fucntion as
然后使用函数作为
echo calculateDistanceBetweenTwoPoints('11.657740','77.766270','11.074820','77.002160','ML',true,5);
Hope it helps
希望能帮助到你
回答by Legy
For exact values do it like that:
对于确切的值，请这样做：
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，我认为应该这样做......
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 JS 实现尝试：http: //www.movable-type.co.uk/scripts/latlong.html
Dare me... I forgot to deg2rad all the values in the circle-functions...
敢不敢...我忘了将圆函数中的所有值 deg2rad...
