解决PHP PDO将数据库整型值映射到类中Enum属性的挑战

// 枚举定义
enum UserType: int
{
    case Master = 1;
    case Admin = 2;
    case Manager = 3;
}

// 用户类定义
class User
{
    private int $id;
    private string $name;
    private UserType $userType; // Enum类型属性
}

// 假设这是你的fetchObject方法
public function fetchObject($sql, array $args = array(), string $class_name = "stdClass"): mixed
{
    $stmt = self::$instance->prepare($sql);
    if(empty($args)){
        $stmt->execute();
    } else{
        $stmt->execute($args);
    }
    $object = $stmt->fetchObject($class_name); // 问题所在
    $stmt->closeCursor();
    return $object;
}

// 调用示例
$user = Database::getInstance()->fetchObject(sql: "SELECT id, name, userType FROM user WHERE id = 1", class_name: User::class);

// Enum定义
enum UserType: int // 必须指定底层类型
{
    case Master = 1;
    case Admin = 2;
    case Manager = 3;
}

// 修改后的User类
class User
{
    private int $id;
    private string $name;
    private UserType $userType; // 声明类型

    public function __construct()
    {
        // 在构造函数中取消设置userType属性，以便PDO调用__set方法
        unset($this->userType);
    }

    // __set魔术方法用于拦截属性赋值
    public function __set($key, $value)
    {
        if ($key === 'userType') {
            // 将整型值转换为UserType枚举实例
            $this->userType = UserType::from($value);
        } else {
            // 处理其他未声明的属性，或抛出错误
            // 最佳实践是避免这种情况，确保所有属性都已声明
            throw new \RuntimeException("Attempt to set unknown or unhandled property: $key");
        }
    }

    // 可以添加getter方法来访问属性
    public function getId(): int { return $this->id; }
    public function getName(): string { return $this->name; }
    public function getUserType(): UserType { return $this->userType; }
}

// 假设你已经有了PDOStatement对象 $stmt
// $stmt = self::$instance->prepare("SELECT id, name, userType FROM user WHERE id = 1");
// $stmt->execute();

// 设置PDO的fetch模式
// PDO::FETCH_CLASS: 创建类的实例
// PDO::FETCH_PROPS_LATE: 先调用构造函数，再设置属性（如果属性不存在，则调用__set）
$stmt->setFetchMode(PDO::FETCH_CLASS | PDO::FETCH_PROPS_LATE, User::class);
$user = $stmt->fetch();

if ($user instanceof User) {
    echo "User ID: " . $user->getId() . "\n";
    echo "User Name: " . $user->getName() . "\n";
    echo "User Type: " . $user->getUserType()->name . " (Value: " . $user->getUserType()->value . ")\n";
} else {
    echo "User not found or fetch failed.\n";
}

// Enum定义保持不变
enum UserType: int
{
    case Master = 1;
    case Admin = 2;
    case Manager = 3;
}

// 修改后的User类（使用构造函数属性提升，PHP 8.0+）
class User
{
    private UserType $userType; // 声明类型

    public function __construct(
        private int $id,
        private string $name,
        int $userType // 构造函数接收整型值
    ) {
        // 在构造函数中将整型值转换为UserType枚举实例
        $this->userType = UserType::from($userType);
    }

    // Getter方法
    public function getId(): int { return $this->id; }
    public function getName(): string { return $this->name; }
    public function getUserType(): UserType { return $this->userType; }
}

// 修改后的fetchObject方法
public function fetchObject($sql, array $args = [], string $class_name = "stdClass"): ?object
{
    $stmt = self::$instance->prepare($sql);
    $stmt->execute($args ?: null); // $args ?: null 处理空数组情况
    $row = $stmt->fetch(PDO::FETCH_ASSOC); // 获取关联数组
    $stmt->closeCursor();

    if ($row) {
        // 使用关联数组解包创建类实例，将数组键值作为命名参数传递给构造函数
        // 要求PHP 8.0+支持命名参数和构造函数属性提升
        return new $class_name(...$row);
    }
    return null;
}

// 调用示例
$user = Database::getInstance()->fetchObject(sql: "SELECT id, name, userType FROM user WHERE id = 1", class_name: User::class);

if ($user instanceof User) {
    echo "User ID: " . $user->getId() . "\n";
    echo "User Name: " . $user->getName() . "\n";
    echo "User Type: " . $user->getUserType()->name . " (Value: " . $user->getUserType()->value . ")\n";
} else {
    echo "User not found or fetch failed.\n";
}