计算从源纬度和经度到目的地的道路距离

Question

我使用了以下代码，但这给了我从源到目的地的空中距离，但我需要道路距离，请帮助我。

这项工作需要这个空中距离，但我需要街道/道路距离。

$center_lat =$_POST['lat'];
//19.159519;//
$center_lng =$_POST['lng'];//72.995782;//
$radius = '3963.0';
$range=$_POST['range'];
//echo $center_lng;
// Set the active mySQL database
$db_selected = mysql_select_db("mapdb", $connection);
// Search the rows in the markers table
$query = sprintf("SELECT name,lat, lng, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM hospitals HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
  mysql_real_escape_string($center_lat),
  mysql_real_escape_string($center_lng),
  mysql_real_escape_string($center_lat),
  mysql_real_escape_string($radius));
$result = mysql_query($query);
if (!$result) {
  die("Invalid query: " . mysql_error());
}
 else
 { 
  echo "Done";
 }
 // header('Content-type: application/json');
//$output = array();
$i=0;
 while($row = mysql_fetch_assoc($result)) {
     if($range > 0 )
      {
        $theta = $center_lng - $row['lng'];
        $distance = (sin(deg2rad($center_lat)) * sin(deg2rad($row['lat']))) + (cos(deg2rad($center_lat)) * cos(deg2rad($row['lat'])) * cos(deg2rad($theta)));   
        $distance = acos($distance);
        $distance = rad2deg($distance);
        $distance = $distance * 60 * 1.1515;
        $distance = $distance *1.609344;
        if($range > round($distance,2))
        {
            $info[$i] = """.$row['lat']."&".$row['lng']."@".$row['name'];
            $i=$i+1;            
        }
        $myTwitterResult1 = array($info);
    }
    else
    {   
        $info[$i] = """.$row['lat']."&".$row['lng']."@".$row['name'];
        $i=$i+1;
        $myTwitterResult1 = array($info);
    }
}
    $myJSONTweets = json_encode($myTwitterResult1);
    echo $myJSONTweets;

Answer 1

如果您正在显示谷歌地图（阅读服务条款）

使用方向Web服务或距离矩阵Web服务

如果您使用Google Maps API v3来显示地图，那么两者在API中都有等效的服务。

Answer 2

此功能可能有助于

function distance($lat1, $lon1, $lat2, $lon2, $unit)
 {
  $angle = $lon1 - $lon2;
  $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) *        cos(deg2rad($lat2)) * cos(deg2rad($angle));
$dist = acos($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;
  }
 }
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";

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