PHP を使用してシンプルなオンラインビデオのアップロードおよび再生システムを実装する方法

CREATE DATABASE IF NOT EXISTS videos;
USE videos;
CREATE TABLE IF NOT EXISTS video_info (
    id INT AUTO_INCREMENT PRIMARY KEY,
    title VARCHAR(255) NOT NULL,
    description TEXT NOT NULL,
    file_name VARCHAR(255) NOT NULL,
    uploaded_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP
);

<?php
if(isset($_POST['upload'])){
    $title = $_POST['title'];
    $description = $_POST['description'];
    $file = $_FILES['video'];

    $file_name = $file['name'];
    $file_tmp = $file['tmp_name'];
    $file_size = $file['size'];
    $file_error = $file['error'];

    if($file_error === UPLOAD_ERR_OK){
        $upload_dir = 'uploads/';
        move_uploaded_file($file_tmp, $upload_dir . $file_name);

        // 将视频信息插入数据库
        $conn = new mysqli('localhost', 'root', 'password', 'videos');
        $conn->query("INSERT INTO video_info(title, description, file_name) VALUES ('$title', '$description', '$file_name')");

        echo "视频上传成功！";
    }else{
        echo "视频上传失败！";
    }
}
?>

<form action="" method="POST" enctype="multipart/form-data">
    <input type="text" name="title" placeholder="请输入视频标题" required><br>
    <textarea name="description" placeholder="请输入视频描述" required></textarea><br>
    <input type="file" name="video" required><br>
    <input type="submit" name="upload" value="上传视频">
</form>

<?php
if(isset($_GET['id'])){
    $id = $_GET['id'];

    $conn = new mysqli('localhost', 'root', 'password', 'videos');
    $result = $conn->query("SELECT * FROM video_info WHERE id = $id");
    $video = $result->fetch_assoc();

    $file_name = $video['file_name'];
    $title = $video['title'];
    $description = $video['description'];
}
?>

<h1><?php echo $title; ?></h1>
<p><?php echo $description; ?></p>

<video controls>
    <source src="uploads/<?php echo $file_name; ?>" type="video/mp4">
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</video>