Question

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”，我想检查一下人的坐标是否在离目标坐标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的？我非常感谢这个解决方案，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!';
    }
}

Answer 1

我有这个功能，它计算点1和2之间的差异。函数返回里程，所以我乘以1.609344将其转换为km。

<?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";
    }
}

Answer 2

您可以使用Great Circle算法执行此操作。 http://en.wikipedia.org/wiki/Great-circle_distance

以下是如何找到距离。

修改

以下是您完整的代码！

<?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; } ?>

当然结果如下：

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

Answer 3

使用毕达哥拉斯依次找到目标与每个人之间的距离，并检查它是否小于2公里。如果x，y是目标和p的坐标，则q是两个人之间距离的坐标：（（x-p）^ 2 +（y-q）^ 2）^（1/2） ^符号的意思是“提升到”的力量。

检查坐标是否在距其他坐标的特定距离内

3 个答案: