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HomeBackend DevelopmentPHP TutorialHow to develop a video sharing website using PHP and CGI

How to use PHP and CGI to develop a video sharing website

With the development of the Internet, video sharing websites have become an important way for people to obtain information and entertainment. If you have a certain understanding of website development, then using PHP and CGI to develop a video sharing website is a good choice. This article will introduce how to use PHP and CGI to implement a simple video sharing website and provide some code examples.

  1. Design website structure
    First of all, we need to design the structure of the website. A typical video sharing website consists of multiple modules, including user registration and login, video upload and sharing, video playback and comments, etc. When designing the website structure, please give full consideration to user experience and interface aesthetics.
  2. Set up database
    Video sharing websites need to use a database to store user information, video information, comments and other data. Use MySQL or other database management systems to create corresponding tables to store this data.

The following is a simple user form example:

CREATE TABLE users (
    id INT PRIMARY KEY AUTO_INCREMENT,
    username VARCHAR(50) NOT NULL,
    password VARCHAR(50) NOT NULL,
    email VARCHAR(50) NOT NULL,
    created_at DATETIME NOT NULL
);
  1. User registration and login
    User registration and login are the basic functions of a video sharing website. User registration and login logic is handled through PHP, and user data is stored in the database.

The following is a simple user registration code example:

<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $username = $_POST['username'];
    $password = $_POST['password'];
    $email = $_POST['email'];

    // 验证用户输入

    // 将用户数据存储到数据库中
    $sql = "INSERT INTO users (username, password, email, created_at)
            VALUES ('$username', '$password', '$email', NOW())";
    // 执行数据库查询

    // 提示用户注册成功
    echo "注册成功!";
}
?>
  1. Video uploading and sharing
    Video uploading and sharing are the core functions of video sharing websites. Use PHP to handle video upload and sharing logic, and save video files to the server.

The following is a simple video upload code example:

<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $title = $_POST['title'];
    $file = $_FILES['video']['tmp_name'];

    // 生成一个唯一的文件名
    $filename = uniqid() . '.mp4';

    // 将视频文件保存到服务器上
    move_uploaded_file($file, 'videos/' . $filename);

    // 将视频信息存储到数据库中
    $sql = "INSERT INTO videos (title, filename, created_at)
            VALUES ('$title', '$filename', NOW())";
    // 执行数据库查询

    // 提示用户上传成功
    echo "上传成功!";
}
?>
  1. Video playback and comments
    Video playback and comment functions can be implemented through CGI. CGI is a method of developing websites that can dynamically generate web content. Video playback and comment functions are implemented through PHP and CGI.

The following is a simple video playback and comment code example:

<?php
// 获取视频ID
$id = $_GET['id'];

// 查询视频信息
$sql = "SELECT * FROM videos WHERE id = $id";
// 执行数据库查询

// 显示视频播放器
echo "<video src='videos/$filename' controls></video>";

// 查询评论信息
$sql = "SELECT * FROM comments WHERE video_id = $id";
// 执行数据库查询

// 显示评论列表
while ($row = mysqli_fetch_assoc($result)) {
    echo "<div>{$row['content']}</div>";
}

// 显示评论表单
echo "<form action='comment.php' method='POST'>
      <textarea name='content'></textarea>
      <input type='hidden' name='video_id' value='$id'>
      <input type='submit' value='评论'>
      </form>";
?>

Through the above steps, we can use PHP and CGI to implement a simple video sharing website. Of course, this is just a starting point, you can expand the functionality of the website and beautify the interface of the website according to your needs. I hope this article is helpful to you, and I wish you success in development!

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