3

好的，所以我在某些坐标上有一个目标，在其他坐标上有一些“人”，我想检查这些人的坐标是否在距离目标坐标 2 公里（2000 米）的范围内。

下面的代码只是为了说明我想要更清楚的内容，问题当然是如何做到这一点？我真的很感激这个问题的解决方案，谢谢！

$person0 = Array('56.34342', '49.324523');
$person1 = Array('57.49544', '47.421524');
$person2 = Array('56.74612', '48.722323');

$target = Array('56.35343', '49.342343');

for (var $i = 0; $i < 4; i$++) {
    CheckIfMatch($person + i$);
}

function CheckIfMatch($person) {
    if($person is within 2km from target) {
        echo 'Match!';
    }
}
4

3 回答 3

4

您可以使用 Great Circle 算法来做到这一点。http://en.wikipedia.org/wiki/Great-circle_distance

以下是如何找到距离。

编辑

这是您执行此操作的完整代码！

<?php
$persons = Array();

$persons[] = Array('52.00951','4.36052');//Delft is more than 2 km from den haag
$persons[] = Array('52.03194','4.31769');//Rijswijk is less than 2 km from den haag
$persons[] = Array('52.07097','4.29945');//A place near den Haag almost 2 streets from center and my favourite coffee shop
$persons[] = Array('52.37022','4.89517');//Amsterdamn is about 60 km *DRIVING* from the hagues according to Gmaps

$target = Array('52.07050', '4.30070');//Den Haag
$i = 0;
foreach($persons as $person){
    $i++;
    $distance = calculate_distance($person, $target);
    if($distance <= 2 ){ 
        echo "Person $i is within 2km with a distance to taget of $distance</br>";
    }else{
        echo "Person $i is <b>not</b> within 2km with a distance to taget of $distance</br>";
    }

}

function calculate_distance($person, $target){

    $lat1 = $person[0];
    $lng1 = $person[1];
    $lat2 = $target[0];
    $lng2 = $target[1];
    $pi = 3.14159;
    $rad = doubleval($pi/180.0);

    $lon1 = doubleval($lng1)*$rad;
    $lat1 = doubleval($lat1)*$rad; 
    $lon2 = doubleval($lng2)*$rad; 
    $lat2 = doubleval($lat2)*$rad; 
    $theta = $lng2 - $lng1; 

    $dist = acos(sin($lat1) * sin($lat2) + cos($lat1) * cos($lat2) * cos($theta)); 

    if ($dist < 0) 
        $dist += $pi;


    $miles = doubleval($dist * 69.1); 
    $inches = doubleval($miles * 63360); 
    $km  =  doubleval($dist * 115.1666667);

    $dist = sprintf( "%.2f",$dist); 
    $miles = sprintf( "%.2f",$miles); 
    $inches = sprintf( "%.2f",$inches); 
    $km = sprintf( "%.2f",$km);
    //Here you can return whatever you please
    return $km;
}

?>

当然还有结果：

Person 1 is not within 2km with a distance to taget of 4.24
Person 2 is within 2km with a distance to taget of 1.21
Person 3 is within 2km with a distance to taget of 0.09
Person 4 is not within 2km with a distance to taget of 41.56
于 2012-12-30T14:47:30.220 回答
3

我有这个函数，它计算点 1 和 2 之间的差。函数返回英里，所以我乘以 1.609344 将其转换为公里。

<?php
function lat_long_dist($lat1, $long1, $lat2, $long2){
    $pi = pi();
    $x  = sin($lat1 * $pi/180) * 
          sin($lat2 * $pi/180) + 
          cos($lat1 * $pi/180) * 
          cos($lat2 * $pi/180) * 
          cos(($long2 * $pi/180) - ($long1 * $pi/180));
    $x  = atan((sqrt(1 - pow($x, 2))) / $x);
    return (1.852 * 60.0 * (($x/$pi) * 180)) / 1.609344;
}

$people   = array();
$people[] = array(56.34342, 49.324523);
$people[] = array(57.49544, 47.421524);
$people[] = array(56.74612, 48.722323);

foreach($people as $person){
    $lat1 = $person[0];
    $lon1 = $person[1];
    $distance = lat_long_dist($lat1, $lon1, 56.35343, 49.342343) * 1.609344;
    if($distance <= 2){
        echo "$i is within 2km!\n";
    }
}
于 2012-12-30T00:30:23.347 回答
0

用毕达哥拉斯依次找出目标与每个人的距离，并检查是否小于2公里。如果 x,y 是目标的坐标，p,q 是其中一个人的坐标，则两者之间的距离为： ((xp)^2 + (yq)^2)^(1/2) ^ 符号意思是“被提升到权力之下”。

于 2012-12-30T00:24:02.933 回答