返回类的构造函数中的值 [英] Returning a value in constructor function of a class

问题描述

到目前为止，我有一个 PHP 类与构造函数

So far I have a PHP class with the constructor

public function __construct ($identifier = NULL)
{
 // Return me.
if ( $identifier != NULL )
{
  $this->emailAddress = $identifier;
  if ($this->loadUser() )
    return $this;      
  else
  {
// registered user requested , but not found ! 
return false;
  }
}

的功能loadUser 是查找特定电子邮件地址的数据库。
当我将标识符设置为某个电子邮件，我确信它不在数据库中;第一个IF被传递，并且去第一个ELSE。这里的构造函数应该返回FALSE;但相反，它返回一个具有所有NULL值的类的对象！

the functionality of loadUser is to look up the database for a particular email address. When i set the identifier to some email that i'm sure it's not in the database; the first IF is get passed, and goes to the first ELSE. here the constructor should return FALSE; but instead, it returns an object of the class with all NULL values !

如何防止这种情况？感谢

how do i prevent this? thanks

编辑：

感谢大家的答案。这是相当快！我看到的OOP方式是抛出一个异常。所以一个抛出一个，我的问题改变，应该怎么做的异常??
php.net的手册很混乱！

thank you all for the answers. that was quite fast ! I see that the OOP way is to throw an Exception. So a throw one, my question changes that what should i do with the exception?? php.net's manual is pretty confusing !

    // Setup the user ( we assume he is a user first. referees, admins are   considered users too )
    try { $him = new user ($_emailAddress);
    } catch (Exception $e_u) { 
      // try the groups database
      try { $him = new group ($_emailAddress); 
      } catch (Exception $e_g) {
          // email address was not in any of them !!  
        }
    }

推荐答案

不获得返回值;

没有重组你已经在做的事情，你可以考虑在这里使用一个异常。

Without restructuring what you are already doing, you may consider using an exception here.

public function __construct ($identifier = NULL)
{
  $this->emailAddress = $identifier;
  $this->loadUser();
}

private function loadUser ()
{
    // try to load the user
    if (/* not able to load user */) {
        throw new Exception('Unable to load user using identifier: ' . $this->identifier);
    }
}

现在，您可以以这种方式创建新用户

Now, you can create a new user in this fashion.

try {
    $user = new User('[e-mail]');
} catch (Exception $e) {
    // unable to create the user using that id, handle the exception
}

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