循環檢查並刪除圖像，如果不存在

Question

我試圖做一個循環，檢查是否存在數據庫的文件名圖像存在，如果不是因爲我有許多重複的文件夾，但文件夾是50GB的所有圖像。我無法檢查每一個。

所以這裏就是我試過

$id = $_GET['id']; 
$sql = "SELECT thumbnail FROM files WHERE id='$id'"; 
$query = $mysqli->query($sql); 
$row = $query->fetch_assoc(); 
$thumb = $row['thumbnail']; 
$records = "../upload/images/"; 
foreach ($records as $record) { 
    if (file_exists('../upload/images/'.$thumb)) {} 
    else { 
     @unlink('../upload/images/'.$thumb); 
    } 
} 

更新

$sql = "SELECT thumbnail FROM filesWHERE id='$id'"; 
$query = $mysqli->query($sql); 
$row = $query->fetchAll(); 

foreach($sql as $search_result) { 
    if(file_exists($search_result['thumbnail'])) { 
     $img_source = ('../upload/images/'.$search_result['thumbnail']); 
    } else { 
     @unlink('../upload/images/'.$search_result); 
    } 
} 

Answer 1

這是我所理解的邏輯。檢查文件夾中是否存在數據庫中不存在的文件，如果存在，請將其刪除。

$directory = "../upload/images/"; 
$images = glob($directory . "*.jpg"); 

foreach($images as $image) 
{ 
    $sql = "SELECT thumbnail FROM files WHERE thumbnail =?"; 
    $stmt = $mysqli->prepare($sql); 
    if($stmt) { 
    $stmt->bind_param('s', $image); 
    $stmt->bind_result($result); 
    $stmt->execute(); 

    $stmt->fetch(); 

    if(!$result) { 
     if(unlink($image)) { 
      echo "Image deleted $image <br>\n"; 
     } 
    } 
    } else { 
     echo "Unable to prepare SQL"; 
    } 

}