Php вывод картинки по id из директории

if (isset($_GET['id'])) { //проверяем, есть ли переменная
  $id = mysqli_real_escape_string($link, $_GET['id']);
  $result = mysqli_query($link, "SELECT image FROM `book` WHERE `id` = '$id'");
  $row = mysqli_fetch_array($result);
    $sql = mysqli_query($link, 'SELECT `id`, `Name`, `Author`, `BookEdition`, `Yearofpublication`, `Jenre`, `NumberofPages` FROM `book`');
    
    }
    if ($sql) {
      while ($result = mysqli_fetch_array($sql)) {
        echo "<tr><td>{$result['id']}</td><td>{$result['Name']}</td><td>{$result['Author']}</td><td>{$result['BookEdition']}</td><td>{$result['Yearofpublication']}</td>
        <td>{$result['Jenre']}</td><td>{$result['NumberofPages']}</td></tr>";
 
      $arr = scandir('Admin/book');
      foreach($arr as $v) {
        continue;
      if(stripos($v,'.jpg')) echo '<a href="book-1.php"><img src=" Admin/book/'.$v.'" width="155" hegiht="220" />';
      }
        
 
    }
    } else {
      echo '<p>Произошла ошибка: ' . mysqli_error($link) . '</p>';
    }

<?php 
$arr = scandir('Admin/book');
foreach($arr as $v) {
if(stripos($v,'.jpg')) echo '<a href="book-1.php"><img src=" Admin/book/'.$v.'" width="155" hegiht="220" />';
}
?>