如何设计一个高效的MySQL表结构来实现视频评论功能？

CREATE TABLE videos (
    id INT AUTO_INCREMENT PRIMARY KEY,
    title VARCHAR(255) NOT NULL,
    description TEXT,
    url VARCHAR(255) NOT NULL,
    created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP
);

CREATE TABLE comments (
    id INT AUTO_INCREMENT PRIMARY KEY,
    video_id INT NOT NULL,
    user_id INT NOT NULL,
    comment TEXT NOT NULL,
    created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
    FOREIGN KEY (video_id) REFERENCES videos(id),
    FOREIGN KEY (user_id) REFERENCES users(id)
);

<?PHP

// 连接到数据库
$conn = new mysqli("localhost", "username", "passWord", "database");

// 检查连接是否成功
if ($conn->connect_error) {
    die("连接失败：" . $conn->connect_error);
}

// 插入视频记录
$sql = "INSERT INTO videos (title, description, url) VALUES ('视频标题', '视频描述', 'Http://example.com/video')";
$conn->query($sql);

// 获取刚插入的视频的ID
$videoId = $conn->insert_id;

// 插入评论记录
$sql = "INSERT INTO comments (video_id, user_id, comment) VALUES ($videoId, 1, '这是一个好视频')";
$conn->query($sql);

// 断开数据库连接
$conn->close();

?>

<?php

// 连接到数据库
$conn = new mysqli("localhost", "username", "password", "database");

// 检查连接是否成功
if ($conn->connect_error) {
    die("连接失败：" . $conn->connect_error);
}

// 获取指定视频的评论列表
$videoId = 1;
$sql = "SELECT * FROM comments WHERE video_id = $videoId";
$result = $conn->query($sql);

// 显示评论列表
if ($result->num_rows > 0) {
    while ($row = $result->fetch_assoc()) {
        echo "用户ID: " . $row["user_id"] . ", 评论内容: " . $row["comment"] . "<br>";
    }
} else {
    echo "暂无评论";
}

// 断开数据库连接
$conn->close();

?>