Question

我如何在类中使用子类？我尝试使用extends，但我认为这不是我想要的。

class Database
{
    protected $link;
    private $host, $database, $username, $password;

    public function __construct($host, $database, $username, $password)
    {
        $this->host = $host;
        $this->database = $database;
        $this->username = $username;
        $this->password = $password;
        $this->connect();
    }

    private function connect()
    {
        $this->link = new mysqli($this->host, $this->username, $this->password, $this->database);

        //GLOBAL $this->link

        if ($this->link->connect_errno)
            die('Connection Error: ' . $this->link->connect_errno);
    }

    public function __sleep()
    {
        return array('host', 'database', 'username', 'password');
    }

    public function __wakeup()
    {
        $this->connect();
    }
}

class Users extends Database{
    public function getAll(){
        $stmt = Database->link->prepare('SELECT id, username, email FROM USERS');
        $stmt->execute();
        $stmt->bind_result($id, $username, $email);
        while($stmt->fetch())
            $row[] = array('id' => $id, 'username' => $username, 'email' => $email);
        $stmt->close();
        return $row;
    }
}

然后在显示内容的文件中......

$db = new Database(HOST, DATABASE, USER, PASSWORD);
$users = $db->Users->getAll();
foreach($users as $user){
    echo $user['username'] . "<br />";
}

我认为我的问题在于$stmt = Database->link->prepare和$users = $db->Users->getAll();

我更像是一个C家伙，所以PHP方法有点让我。有人可以准确地描述我需要做什么才能获得正确的子类化方法吗？

Answer 1

我认为我的问题在于$stmt = Database->link->prepare和$users = $db->Users->getAll()

你做对了。

当Users扩展Database时，它可以访问受保护的属性$link，因此，您只需通过$this使用它，例如

$stmt = $this->link->prepare(...)

现在，每次实例化Database类或其后代时，您都不希望创建新的MySQL连接。相反，您应该将mysqli依赖项传递给构造函数，例如

abstract class Database {
    protected $link;

    public function __construct(mysqli $link) {
        $this->link = $link;
    }
}

class Users extends Database {
    public function getAll() {
        $stmt = $this->link->prepare(...);
        // and so on
    }
}

$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$usersDao = new Users($mysqli);
$users = $usersDao->getAll();
foreach($users as $user){
    echo $user['username'] . "<br />";
}

正在使用的PHP子类

1 个答案: