PHP와 SQLite: 이미지와 멀티미디어 파일을 처리하는 방법

$ sqlite3 media.db
SQLite version 3.31.1 20...
Enter ".help" for usage hints.

sqlite> CREATE TABLE media (
   id INTEGER PRIMARY KEY,
   filename TEXT NOT NULL,
   type TEXT NOT NULL,
   filepath TEXT NOT NULL
);
sqlite> .quit

<form action="upload.php" method="post" enctype="multipart/form-data">
   <input type="file" name="file">
   <input type="submit" value="Upload">
</form>

<?php
if(isset($_FILES['file'])) {
   $file = $_FILES['file'];
   $filename = $file['name'];
   $filetype = $file['type'];
   $filesize = $file['size'];
   $tmpname = $file['tmp_name'];
   
   $filepath = 'uploads/'. $filename;
   
   move_uploaded_file($tmpname, $filepath);
   
   // 插入文件信息到SQLite数据库
   $db = new SQLite3('media.db');
   $stmt = $db->prepare("INSERT INTO media (filename, type, filepath) VALUES (:filename, :type, :filepath)");
   $stmt->bindValue(':filename', $filename, SQLITE3_TEXT);
   $stmt->bindValue(':type', $filetype, SQLITE3_TEXT);
   $stmt->bindValue(':filepath', $filepath, SQLITE3_TEXT);
   $stmt->execute();
   
   echo "文件上传成功！";
}
?>

<?php
$db = new SQLite3('media.db');
$stmt = $db->prepare("SELECT * FROM media");
$result = $stmt->execute();

while($row = $result->fetchArray(SQLITE3_ASSOC)) {
   $filepath = $row['filepath'];
   $filetype = $row['type'];
   
   if($filetype === 'image/jpeg' || $filetype === 'image/png') {
      echo "<img src='$filepath' alt=''>";
   }
   elseif($filetype === 'audio/mpeg') {
      echo "<audio controls><source src='$filepath' type='audio/mpeg'></audio>";
   }
   elseif($filetype === 'video/mp4') {
      echo "<video controls><source src='$filepath' type='video/mp4'></video>";
   }
}
?>