無法使用PDO方法時，使用數據庫類

Question

嘿傢伙我這樣做再次錯誤我敢肯定，但我想實例化PDO數據庫 處理程序從我的班級數據庫從文件class.database.php在class.admin.php的AdminSession 裏面，有些東西對我的依賴注入有些詭異，而且它不是 ，它允許我正確地使用PDO的方法;像fetch（），prepare（）etcetra。

的class.database.php文件

class Database 
{ 
    public $db; // handle of the db connection 

    private static $dsn="mysql:host=server2.com;dbname=database"; 
    private static $user="user"; 
    private static $pass="pass"; 
    private static $instance; 

    public function __construct() 
    { 
     $this->db = new PDO(self::$dsn,self::$user,self::$pass,$self::$opts); 
     $this->db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
     $this->db->setAttribute(PDO::MYSQL_ATTR_INIT_COMMAND, "SET NAMES 'utf8'"); 

     switch($_SERVER['ENVIRONMENT']) { 
      case 'staging': 
       self::$dsn="mysql:host=server1.com;dbname=database"; 
       self::$user="user"; 
       self::$pass="pass"; 
       break; 
      default: 
       self::$dsn="mysql:host=server2.com;dbname=database"; 
       self::$user="user"; 
       self::$pass="pass"; 
     } 

    } 

    public static function getInstance() 
    { 
     if(!isset(self::$instance)) 
     { 
      $object= __CLASS__; 
      self::$instance=new $object; 
     } 
     return self::$instance; 
    } 

} 

，這裏是最頂層的我class.admin.php的，那就是拋出一個錯誤的方法。 現在我越來越

PHP致命錯誤的錯誤：調用未定義的方法行230

如果我使用$this->db-prepare($sql)

或

PHP致命錯誤：調用到成員函數prepare（）在非對象行上230

如果我使用$db-prepare($sql)

require('library/class.database.php'); 

class AdminSession { 
    static $abs_path; 

    public function __construct(Database $db) { 
     session_start(); 

     self::$abs_path = dirname(dirname(__FILE__)); 
     if($_SERVER['REQUEST_METHOD'] == 'POST') { 

      $this->post = $_POST; // filter_input_array(INPUT_POST, FILTER_SANITIZE_STRING); 

      if(get_magic_quotes_gpc()) { 
       //get rid of magic quotes and slashes if present 
       array_walk_recursive($this->post, array($this, 'stripslash_gpc')); 
      } 
     } 

     $this->get = $_GET; // filter_input_array(INPUT_GET, FILTER_SANITIZE_STRING); 
     array_walk_recursive($this->get, array($this, 'urldecode')); 
    } 

// other methods 

    private function checkDB($username, $password) { 
     $sql = "SELECT * FROM users WHERE username=:username"; 
     try { 

      $db    = Database::getInstance(); 
      $stmt   = $db->prepare($sql); 

       $stmt->bindParam("username", $username); 
       $stmt->execute(); 

      $user   = $stmt->fetchAll(PDO::FETCH_OBJ); 

      $db = null; 

      if($user) { 
       //general return 
       if(is_object($user[0]) && md5($user[0]->password) == $password) { 
        return true; 
       } else { 
        return false; 
       } 
      } else { 
       return false; 
      } 

     } catch(PDOException $e) { 
      echo '{"error":{"text":'. $e->getMessage() .'}}'; 
     } 

    } 

} 

Answer 1

你不能把邏輯放到你的類定義中。相反，請確定構造函數中這些變量的值。該開關將在一個方法中工作，但不能在定義成員時使用。

編輯：我真的覺得很愚蠢。該連接在switch語句之前完成。我不知道它會解決第二組問題......但它現在對於原來的問題會表現得很好。

class Database 
{ 
    public $db; // handle of the db connection 

    private static $opts = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8'); 
    private static $dsn="mysql:host=server2.com;dbname=database"; 
    private static $user="user"; 
    private static $pass="pass"; 
    private static $instance; 

    public function __construct() 
    { 
     switch($_SERVER['ENVIRONMENT']) { 
      case 'staging': 
       self::$dsn="mysql:host=server1.com;dbname=database"; 
       self::$user="user"; 
       self::$pass="pass"; 
       break; 
      default: 
       self::$dsn="mysql:host=server2.com;dbname=database"; 
       self::$user="user"; 
       self::$pass="pass"; 
     } 

     $this->db = new PDO(self::$dsn,self::$user,self::$pass,$self::$opts); 
     $this->db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
    } 

    public static function getInstance() 
    { 
     if(!isset(self::$instance)) 
     { 
      $object= __CLASS__; 
      self::$instance=new $object; 
     } 
     return self::$instance; 
    } 

}