如何使用PHP检查某个坐标是否落入另一个坐标半径 [英] How to check if a certain coordinates fall to another coordinates radius using PHP only

问题描述

我看过这么多的函数，但它恰好只适用于MySQL或Postgresql。我想要PHP的等价逻辑。我正在做一些比较，
就像我有这些数据是在创建时产生的。

 纬度：56.130366 
 Long：-106.34677099999 
  

稍后，我想检查这个坐标是否会落入

>纬度：57.223366
长度：-106.34675644699
半径：100000（米）

解决方案

感谢您的帮助。下面是一个示例函数，它使用两组经度和纬度坐标并返回两者之间的距离。

  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; 
 
返回$ d; 
} 
 
 $ distance = getDistance（56.130366，-106.34677099999，57.223366，-106.34675644699）; 
 if（$ distance <100）{
 echo100公里范围内; 
} else {
 echo100公里半径以外; 
} 
  

I have seen so many functions but it happens to work only for MySQL or Postgresql. I want the equivalent logic for PHP. I'm doing some comparisons, like I have this data that were being produced when created.

Lat: 56.130366
Long: -106.34677099999

Later on, I want to check if this coordinates will fall within a radius of another coordinates then returns true, otherwise false.

Lat: 57.223366
Long: -106.34675644699
radius: 100000 ( meters )

Thanks in advance!

解决方案

Thanks for the help. Below is an example function that takes two sets of longitude and latitude co-ordinates and returns the distance between the two.

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;  
}

$distance = getDistance( 56.130366, -106.34677099999, 57.223366, -106.34675644699 );
if( $distance < 100 ) {
    echo "Within 100 kilometer radius";
} else {
    echo "Outside 100 kilometer radius";
}

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