php - 如何在另一个类中使用一个类？

Question

我有一个类，它有一些关于帐户管理的功能，我还有一个验证电子邮件地址、用户名等的类。

如何在帐户类中使用验证类？我如何轻松地包含它？

Answer 1

您可以实例化验证类并将其作为参数传递给帐户类，并将其设置为帐户构造方法中帐户类中的属性吗？

Answer 2

如果一个类是专门为验证目的而设计的，而另一个类是为包含特定于实例的信息而设计的，那么您可以采用三种不同的方法。

1）静态引用验证器（推荐）：

<?php
class Validation {
  public static function validateEmail($email) {
    return filter_var($email, FILTER_VALIDATE_EMAIL);
  }
}

class AccountManagement {
  public $email;
  public function __construct($email) {
    $this->email = $email;

    // Validate the e-mail. If not valid, an exception is thrown.
    if(!Validation::validateEmail($this->email)) {
      throw new InvalidArgumentException('$email argument supplied must contain a valid e-mail address');
    }
  }
}

2）扩展您的实例类以从您的验证类继承

<?php
class Validation {
  public function validateEmail($email) {
    return filter_var($email, FILTER_VALIDATE_EMAIL);
  }
}

class AccountManagement extends Validation {
  public $email;
  public function __construct($email) {
    $this->email = $email;

    // Validate the e-mail. If not valid, an exception is thrown.
    if(!$this->validateEmail($this->email)) {
      throw new InvalidArgumentException('$email argument supplied must contain a valid e-mail address');
    }
  }
}

3）从您的实例类中实例化您的验证类（不推荐）：

<?php
class Validation {
  public function validateEmail($email) {
    return filter_var($email, FILTER_VALIDATE_EMAIL);
  }
}

class AccountManagement extends Validation {
  public $email;
  public function __construct($email) {
    $this->email = $email;

    $validator = new Validation;

    // Validate the e-mail. If not valid, an exception is thrown.
    if(!$validator->validateEmail($this->email)) {
      throw new InvalidArgumentException('$email argument supplied must contain a valid e-mail address');
    }
  }
}

测试你的实现

一旦您选择了最适合您需求的方法（无论您选择哪一种），您就可以使用以下代码对其进行测试：

// Valid, should not throw an exception and should print success.
try {
  $account = New AccountManagement('me@myself.com');
  print "AccountManagement object successfully instantiated.<br />\r\n";
} catch(Exception $e) {
  print 'Error: Encountered ' . $e;
}

// Invalid, should throw an InvalidArgumentException exception
try {
  $account = New AccountManagement('myself.com');
  print "AccountManagement object successfully instantiated.<br />\r\n";
} catch(Exception $e) {
  print 'Error: Encountered ' . $e;
}

奖金示例：

有时，您可能希望专门捕获验证错误，并让某些东西处理可能遇到的任何其他异常。在这种情况下，我们可以为我们的验证器创建一个特殊的异常，并允许验证器是抛出异常的那个：

<?php
class ValidationException extends Exception {
  // Will use default Exception behavior.
}

class Validation {
  public static function validateEmail($email) {
    if(!filter_var($email, FILTER_VALIDATE_EMAIL)) {
      throw new ValidationException('E-mail address supplied is not valid');
    }
  }
}

class AccountManagement {
  public $email;
  public function __construct($email) {
    $this->email = $email;

    // Validate the e-mail.
    Validation::validateEmail($this->email);
  }
}

// Valid, should not throw an exception and should print success.
try {
  $account = New AccountManagement('me@myself.com');
  print "AccountManagement object successfully instantiated.<br />\r\n";
} catch(Exception $e) {
  print 'Error: Encountered ' . $e;
}

// Invalid, should throw an ValidationException exception
try {
  $account = New AccountManagement('myself.com');
  print "AccountManagement object successfully instantiated.<br />\r\n";
} catch(ValidationException $e) {
  print 'Error: Encountered ' . $e;
}

Answer 3

您可以使用方法创建第三类“AccountsManagementValidator”

validate(AccountManagement): void
hasErrors(): boolean
getErrors(): List<Error>

有很多不同的实现，但我描述的是 PlayFramework 验证器实现。

当然，您的“验证器”类会重用您的常用方法来验证电子邮件、用户等......