php php的Haversine公式
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14750275/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Haversine formula with php
提问by martinstoeckli
I want to use this formula with php. I have a database with some values of latitute and longitude saved.
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。
I want to find, with a certain value of latitude and longitude in input, all the distances (in km) from this point with each point in the database. To do this, I used the formula on googlemaps api:
我想在输入的纬度和经度的特定值下找到数据库中每个点与该点的所有距离（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 ) ) ) )
Of course using that in php I replaced radians with deg2rad.The values 37,-122 are my values of input and lat,lng are my values in the database.
当然，在 php 中使用它我将弧度替换为deg2rad。值 37,-122 是我的输入值，lat,lng 是我在数据库中的值。
Below there is my code. The problem is that there is something wrong but I don't understand what. The value of distance is of course wrong.
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))))) );
}
For values for example: $lat= 41.9133741000 $lng= 12.5203944000
对于例如值： $lat= 41.9133741000 $lng= 12.5203944000
I have the output of distance="4826.9341106926"
我有 distance="4826.9341106926" 的输出
回答by martinstoeckli
The formula you used, seems to be the arccosineinstead of the haversineformula. The haversine formula is indeed more appropriate to calculate the distance on a sphere, because it is less prone to rounding errors.
您使用的公式似乎是反余弦而不是半正弦公式。半正弦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;
}
P.S. I couldn't find an error in your code, so is it just a typo that you wrote $lat= 41.9133741000 $lat= 12.5203944000? Maybe you just calculated with $lat=12.5203944000 and $long=0 because you overwrote your $lat variable.
PS 我在你的代码中找不到错误，所以这只是你写的错字$lat= 41.9133741000 $lat= 12.5203944000吗？也许你只是用 $lat=12.5203944000 和 $long=0 计算，因为你覆盖了你的 $lat 变量。
Edit:
编辑：
Tested the code and it returned a correct result:
测试了代码并返回了正确的结果：
$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
回答by Tayyab Hussain
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 $distance; // in km
}
回答by Naryl
from this link:
从这个链接：
function getDistance($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;
}
As you can see there are many differences between this as your code. I don't know if you have either a different approach to the formula or maybe some step when converting to PHP went wrong, but the above formula should work.
正如您所看到的，这与您的代码之间存在许多差异。我不知道您是否对公式有不同的方法，或者在转换为 PHP 时的某些步骤出错了，但上面的公式应该可以工作。
回答by CyberBrain
I calculate distances straight inside queries, using the following stored procedure:
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|
You just execute the above as an SQl statement from within phpMyAdmin to create the procedure. Just notice the ending |, so in your SQL input window, choose for the | sign as limiter.
您只需在 phpMyAdmin 中将上述内容作为 SQL 语句执行即可创建过程。请注意结尾 |，因此在您的 SQL 输入窗口中，选择 | 作为限制者签名。
Then in a query, call it like this:
然后在查询中，像这样调用它：
$sql = "
SELECT `locations`.`name`, GEODIST(`locations`.`lat`, `locations`.`lon`, " . $lat_to_calculate . ", " . $lon_to_calculate . ") AS `distance`
FROM `locations` ";
I found this to be a lot faster than calculating it in PHP after the query has been run.
我发现这比在查询运行后用 PHP 计算要快得多。
回答by Sandip Bhoi
I making the class of haversign which having the static fuction getDistance having the four paramters and it return the distance from the bot location points
我创建了具有四个参数的静态函数 getDistance 的 hasrsign 类，它返回与机器人位置点的距离
class HaverSign {
public static function getDistance($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;
}
}
above class is stored at the root directory and root directory contains classes folder Call this by using the following way in any php page
上面的class存放在根目录下，根目录下包含classes文件夹在任意php页面中使用如下方式调用
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;
Output is as follow
输出如下
4.7676529976827
