Ошибка второго запроса MySQL MySQL

Question

Кажется, мой второй запрос MySQL не работает по неизвестным причинам. У меня нет проблем при подключении, только когда вы получите результат из $ dbLink-> query();

$photos = getPhotos(getID($_COOKIE[$uname_cookie], $_COOKIE[$passwd_cookie])); 

function getID($uName, $passwd) { 
    // Just in case, for modularity :P 
    include 'misc.php'; 
    $query = "SELECT ID FROM Users WHERE UserName = '" . $uName . "' AND Password = '" . $passwd . "';"; 
    $dbLink = new mysqli($host, $user, $pass, $dbname); 
    if ($dbLink->connect_errno) { 
     echo "Failed to connect to database: " . $dbLink->connect_error; 
     return null; 
    } else { 
     $results = $dbLink->query($query); 
     print_r($results->fetch_array()); 
     $rows = $results->fetch_array(); 
     return $rows["ID"]; 
    } 
} 

function getPhotos($id) { 
    // Just in case, for modularity :P 
    include 'misc.php'; 
    $query = "SELECT Location FROM Photos WHERE UserID = " . $id . ";"; 
    $dbLink = new mysqli($host, $user, $pass, $dbname); 
    if ($dbLink->connect_errno) { 
     echo "Failed to connect to database: " . $dbLink->connect_error; 
     return null; 
    } else { 
     $results = $dbLink->query($query); 
     print_r($results->fetch_array()); 
     $rows = $results->fetch_array(); 
     return $rows["ID"]; 
    } 
} 

EDIT: Есть пропусканием вокруг $ DBLink к функциям IE getPhotos ($ ID, $ DBLink)

Answer 1

$query = "SELECT Location FROM Photos WHERE UserID = '" . $id . "';";