从另一个类访问类变量


Accessing a class variable from another class

我对PHP OO概念相对较新。我一直在搞砸以下内容,

<?php
require('class_database.php');
require('class_session.php');
class something {  
  $database = NULL;
  $session = NULL;
  function __construct() {
    $this->database = new database();
    $this->session = new session();
  }
  function login() {
    $this->session->doLogin();
  }
}
?>
in another script
$something = new something();
$something->login();

在此示例中,$database有一个构造函数,该构造函数创建一个包含 MySQLi 连接的受保护变量。它还有一个函数叫做 query() .

如果需要运行$database->query $session,我该怎么做?创建一个新的数据库类会非常浪费,但我似乎无法从$session访问$database

我是否给自己制造了一个本应避免的大麻烦?我希望会话提供登录验证,这需要访问数据库以检查凭据。

感谢您的任何帮助

有关此主题的详细信息,请参阅 http://en.wikipedia.org/wiki/Dependency_Injection 和 http://martinfowler.com/articles/injection.html。

<?php
class something {  
  require('class_database.php');
  require('class_session.php');
  $database = new database();
  // dependency injection either here in constructor or in the function that needs it below
  $session = new session($database);

  $session->doSomething(); //This method wants to access $database->query();
}
?>

更新(OP更改了有问题的代码)

<?php
require('class_database.php');
require('class_session.php');
class something
{  
  protected $session = NULL;
  public function __construct($session) {
    $this->session = $session;
  }
  public function login() {
    $this->session->doLogin();
  }
}
$database = new database();
$session = new session($database);
$something = new something($session);
$something->login();
?>

如何使会话类的构造函数采用作为数据库实例的参数?

$database = new database();
$session = new session($database);

然后在会话构造函数中,将其设置为对象的属性并使用它:

private var $_database;
public function __construct($database) {
   $this->_database = $database;
}
public function doSomething() {
    $this->_database->query();
}