我使用了以下代码,但这给了我从源头到目的地的空中距离,但我需要道路距离,请帮帮我。
这项工作,但这个空中距离,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;