i use mysql xampp for database and i use that database for save my thumbnails.
this my code for the dashboard buat idk for the error or maybe just require something?
please give tell me
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Dashboard</title>
<link rel="stylesheet" href="style_dashboard.css">
</head>
<body>
<!-- Header -->
<header>
<h1>Perpustakaan Online</h1>
<nav>
<ul>
<li><a href="./dashboard.php">Home</a></li>
<li><a href="profile.php">Profile</a></li>
<li><a href="../logout.php">Logout</a></li>
</ul>
</nav>
</header>
<!-- Main Content -->
<main>
<h2>Dashboard Buku</h2>
<form method="get">
<input type="text" name="search" placeholder="Search...">
<input type="submit" value="Search">
</form>
<ul class="book-list">
<?php
// Membuat koneksi ke database
$host = 'localhost';
$user = 'root';
$password = '';
$database = 'library_db';
$conn = mysqli_connect($host, $user, $password, $database);
// Memeriksa koneksi
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();
}
// Membuat query untuk mengambil data buku
$query = "SELECT * FROM book";
// Menambahkan kriteria pencarian jika ada
if (isset($_GET['search'])) {
$search = mysqli_real_escape_string($conn, $_GET['search']);
$query .= " WHERE title LIKE '%$search%' OR author LIKE '%$search%'";
}
// Menjalankan query
$result = mysqli_query($conn, $query);
// Menampilkan daftar buku
while ($row = mysqli_fetch_assoc($result)) {
echo "<li>";
echo "<img src='" . $row['thumbnail'] . "' alt='" . $row['title'] . "'>";
echo "<h3>" . $row['title'] . "</h3>";
echo "<p>Penulis: " . $row['author'] . "</p>";
echo "</li>";
}
// Menutup koneksi
mysqli_close($conn);
?>
</ul>
</main>
<!-- Footer -->
<footer>
<p>© 2023 Perpustakaan Online. All rights reserved.</p>
</footer>
</body>
</html>
i think that from the name of picture(thumbnail) or from the database or the code and maybe from th image format