SQL或PHP错误？距离计算

Question

我的文件有问题。我正在制作Travian Clone脚本，我们走得很远。现在我们决定添加人工制品进入游戏。

目标是展示最接近文物到目前该村我们中的代码是：

function getDistance($coorx1, $coory1, $coorx2, $coory2) { 
      $max = 2 * WORLD_MAX + 1; 
      $x1 = intval($coorx1); 
      $y1 = intval($coory1); 
      $x2 = intval($coorx2); 
      $y2 = intval($coory2); 
      $distanceX = min(abs($x2 - $x1), $max - abs($x2 - $x1)); 
      $distanceY = min(abs($y2 - $y1), $max - abs($y2 - $y1)); 
      $dist = sqrt(pow($distanceX, 2) + pow($distanceY, 2)); 

      return round($dist, 1); 
     } 


     unset($reqlvl); 
     unset($effect); 
     $arts = mysql_query("SELECT * FROM ".TB_PREFIX."artefacts WHERE id > 0"); 
     $rows = array(); 
     while($row = mysql_fetch_array($arts)) { 
      $query = mysql_query('SELECT * FROM `' . TB_PREFIX . 'wdata` WHERE `id` = ' . $row['vref']); 
      $coor2 = mysql_fetch_assoc($query); 

      $wref = $village->wid; 
      $coor = $database->getCoor($wref); 
      $dist = getDistance($coor['x'], $coor['y'], $coor2['x'], $coor2['y']); 

      $rows[$dist] = $row; 

     } 
     ksort($rows, SORT_ASC); 
     foreach($rows as $row) { 
      echo '<tr>'; 
      echo '<td class="icon"><img class="artefact_icon_'.$row['type'].'" src="img/x.gif" alt="" title=""></td>'; 
      echo '<td class="nam">'; 
      echo '<a href="build.php?id='.$id.'">'.$row['name'].'</a> <span class="bon">'.$row['effect'].'</span>'; 
      echo '<div class="info">'; 
      if($row['size'] == 1){ 
        $reqlvl = 10; 
        $effect = "village"; 
       }elseif($row['size'] == 2 OR $row['size'] == 3){ 
        $reqlvl = 20; 
        $effect = "account"; 
       } 
      echo '<div class="info">Treasury <b>'.$reqlvl.'</b>, Effect <b>'.$effect.'</b>'; 
      echo '</div></td><td class="pla"><a href="karte.php?d='.$row['vref'].'&c='.$generator->getMapCheck($row['vref']).'">'.$database->getUserField($row['owner'],"username",0).'</a></td>'; 
      echo '<td class="dist">'.getDistance($coor['x'], $coor['y'], $coor2['x'], $coor2['y']).'</td>'; 
      echo '</tr>'; 
     } 
?> 

但代码似乎是错误的，因为它显示了所有相同的距离。 14.8给我。我知道我也许有不好的解释，但你可能会明白我需要什么。

Answer 1

我不能与你当前的代码帮我怕，但你可以尝试使用Haversine Formula代替：

// Where: 
// $l1 ==> latitude1 
// $o1 ==> longitude1 
// $l2 ==> latitude2 
// $o2 ==> longitude2 
function haversine ($l1, $o1, $l2, $o2) { 
    $l1 = deg2rad ($l1); 
    $sinl1 = sin ($l1); 
    $l2 = deg2rad ($l2); 
    $o1 = deg2rad ($o1); 
    $o2 = deg2rad ($o2); 

    $distance = (7926 - 26 * $sinl1) * asin (min (1, 0.707106781186548 * sqrt ((1 - (sin ($l2) * $sinl1) - cos ($l1) * cos ($l2) * cos ($o2 - $o1))))); 

    return round($distance, 2); 
} 

幸得this post上go4expert.com，我用这个功能在过去并发现它工作得很好。