php是不匹配ios lon和拉特與mysql的lon和lat正確

Question

我想從ios（這是工作正常）得到輸出的經緯度和座標，發送到PHP與MySQL查詢和有PHP發送一個xml文檔返回給ios（這一步不工作，因爲它沒有帶回該位置的mysql條目），然後解析它在iOS UItableview（這也工作得很好）。我試圖讓這與XML工作，因爲我已經得到了一個簡單的XML腳本已經在它上面運行。但可能是由於缺乏經驗的PHP錯誤，我無法獲得這個PHP腳本的工作！我在我的PHP腳本中做錯了什麼？謝謝！哦，而且，在mysql中的數據類別由「lon」和「lat」和「name」（爲附近的朋友或家人的名字）組成！而如果有人想知道，這是一個較早的腳本的演進版本（這也產生了相同的結果）：php query for iOS latitude and longitude not searching for nearby mysql lat and lon with a xml output

<?php 
    define('LATMILES', 1/69); 
    define('LONMILES', 1/53); 
    if (isset($_GET['lat'])) { $lat = (float)$_GET['lat']; } //Recieve ios input from: NSString *urlString = [NSString stringWithFormat:@"http://www.mysite.com/loc.php?lat=%g&lon=%g&radius=100&q=%@", latitude, longitude, searchBar.text?searchBar.text:@""]; 
    if (isset($_GET['lon'])) { $lon = (float)$_GET['lon']; } //Recieve ios input from: NSString *urlString = [NSString stringWithFormat:@"http://www.mysite.com/loc.php?lat=%g&lon=%g&radius=100&q=%@", latitude, longitude, searchBar.text?searchBar.text:@""]; 
    if (isset($_GET['radius'])) { $radius = (float)$_GET['radius']; } //Recieve ios input from: NSString *urlString = [NSString stringWithFormat:@"http://www.mysite.com/loc.php?lat=%g&lon=%g&radius=100&q=%@", latitude, longitude, searchBar.text?searchBar.text:@""]; 
    $minlat = $lat - ($radius * LATMILES); 
    $minlon = $lon - ($radius * LONMILES); 
    $maxlat = $lat + ($radius * LATMILES); 
    $maxlon = $lon + ($radius * LONMILES); 
    $dbh = new PDO('(censored private information'); 
    $sql = 'SELECT lat, lon, name FROM locations WHERE lat >= ? AND lat <= ? AND lon >= ? AND lon <= ?'; 
    $params = array($minlat, $maxlat, $minlon, $maxlon); 
    if (isset($_GET['q'])) { 
     $sql .= " AND name LIKE ?"; 
     $params []= '%'.$_GET['q'].'%'; 
    } 
    $q = $dbh->prepare($sql); 
    $q->execute($params); 
    $doc = new DOMDocument(); 
    $r = $doc->createElement("locations"); 
    $doc->appendChild($r); 
    foreach ($q->fetchAll() as $row) { 
     $dlat = ((float)$row['lat'] - $lat)/LATMILES; 
     $dlon = ((float)$row['lon'] - $lon)/LONMILES; 
     $d = sqrt(($dlat * $dlat) + ($dlon * $dlon)); 
     if ($d <= $radius) { 
     $e = $doc->createElement("location"); 
     $e->setAttribute('lat', $row['lat']); 
     $e->setAttribute('lon', $row['lon']); 
     $e->setAttribute('name', $row['name']); 
     $r->appendChild($e); 
     } 
    } 
    print $doc->saveXML(); 
?> 

Answer 1

你的代碼是一個過於複雜。首先，尋找位置的最佳解決方案是使用Haversine_formula。

我已簡化您的代碼以使用它。

<?php 
if (isset($_GET['lat'])){ 
    $lat = (float)$_GET['lat']; 
} 
if (isset($_GET['lon'])){ 
    $lon = (float)$_GET['lon']; 
} 
if (isset($_GET['radius'])){ 
    $radius = (float)$_GET['radius']; 
} 
if (isset($_GET['q'])){ 
    $name = $_GET['q']; 
} 

$dbh = new PDO('(censored private information'); 
// 
$sql = "SELECT name, lat, lon, (3959 * acos(cos(radians('%s')) * cos(radians(lat)) * cos(radians(lon) - radians('%s')) + sin(radians('%s')) * sin(radians(lon)))) AS distance FROM location WHERE `name` LIKE '%s' HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20", 
mysql_real_escape_string($lat), 
mysql_real_escape_string($lon), 
mysql_real_escape_string($lat), 
mysql_real_escape_string($name), 
mysql_real_escape_string($radius)); 

$q = $dbh->prepare($sql); 
$q->execute($params); 
$doc = new DOMDocument(); 
$r = $doc->createElement("locations"); 
$doc->appendChild($r); 
foreach ($q->fetchAll() as $row) { 
    //$dlat = ((float)$row['lat'] - $lat)/LATMILES; 
    //$dlon = ((float)$row['lon'] - $lon)/LONMILES; 
    //]$d = sqrt(($dlat * $dlat) + ($dlon * $dlon)); 
    //if ($d <= $radius) { 
    $e = $doc->createElement("location"); 
    $e->setAttribute('lat', $row['lat']); 
    $e->setAttribute('lon', $row['lon']); 
    $e->setAttribute('name', $row['name']); 
    $r->appendChild($e); 
    //} 
} 
echo $doc->saveXML(); 
?> 

我已經註釋掉了一些你的代碼，因爲我不認爲這是必需的。

我也改變了print $doc->saveXML();到echo $doc->saveXML();

我也想嘗試硬編碼參數，即$lat =55.00;等，以確保獲得所需的輸出。

上述公式使用公式改爲公里使用6371代替3959