php - 为什么不调用父构造函数？

Question

我将 parent::__construct(); 添加到表和书签的构造函数中，以便让这段代码正常工作。为什么不自动调用它们？

如果我创建一个书签类型的对象 $obj_ref_bo = new bookmark(); 不应该书签也从它的每个父类(除了抽象类)创建对象。

调用链是

bookmark->table->database(abstract)->single_connect

    /*single_connect*/

class single_connect
  {
  protected static $_db_pointer = NULL;
  private function __construct()
    {
    $this->_db_pointer = mysql_connect(DB_HOST, DB_USER, DB_PASS);
    mysql_select_db(DB_DATABASE);
    }
  public static function get_connection()
    {
    if(self::$_db_pointer == NULL)
      {
      return new self();
      } 
    else
      {
      echo "Error:only one connection";
      }
    }
  }

/*database*/

abstract class database
  {
  protected function __construct()
    {
    single_connect::get_connection();
    }
  static protected function query($query)
    {
    $result = mysql_query($query) or die(mysql_error());
    return $result;
    }
  }

/*table*/

class table extends database
  {
  public $_protected_arr=array();
  protected function __construct()
    {
    parent::__construct();
    $this->protect();
    }
  protected function protect()
    {
    foreach($_POST as $key => $value)
      {
      $this->_protected_arr[$key] = mysql_real_escape_string($value);
      }
    }
  }

/*bookmark*/

class bookmark extends table 
  {
  function __construct()
    {
    parent::__construct();
    $this->start();
    }   
  function start()
    {
    if(this->test())
      {
      this->insert();
      }
    else
      {
      return 1;
      }
    }
  function test()
    {
    if(this->test_empty())
      {
      return 1;
      }
    else
      {
      return 0;
      }
    }
  function test_empty()
    {
    if(text::test_empty($this->_protected_arr)) 
      {
      return 1;
      }
    else
      {
      return 0;
      }
    }
  function insert()
    {
    $url =  $this->_protected_arr['url'];
    $title =  $this->_protected_arr['title'];
    $email = $_SESSION['email'];
    database::query("INSERT INTO bo VALUES ('$title', '$url', '$email')");
    }
  }

Answer 1

should not bookmark also create objects from each of its parent classes

这完全是您的选择，语言中没有要求调用父方法。

正如 PHP 手册简洁地指出的那样:

Note: Parent constructors are not called implicitly if the child class defines a constructor. In order to run a parent constructor, a call to parent::__construct() within the child constructor is required.

— http://php.net/oop5.decon