Question

我刚开始学习OOP PHP，我正在尝试创建类，它将连接到我的数据库。

代码：

  class DB_CONNECT
    {
    private $host ;
    private $dbName ;
    private $userName ;
    private $password;
    private $db;

    public function __construct($host,$dbName,$userName,$password){

        $this->host = $host;
        $this->dbName = $dbName;
        $this->userName = $userName;
        $this->password = $password;

        try {

             $this->db = new PDO('mysql:host='.$this->host.';dbname='.$this->dbName.';charset=utf8',$this->userName,$this->password);
             $this->db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
             return $this->db;
        } catch (Exception $e) {

            ECHO $e->getMessage();
        }

    }   

}

 $db = new DB_CONNECT("localhost", "oopcms","viktor","viktor");


 function select($db){

    $query = $db->prepare("SELECT * FROM `test`");
    $query->execute();
    $row = $query->fetchAll(PDO::FETCH_ASSOC);
    return $row;
 }

 $x = select($db);
 var_dump($x);

但是我收到了这个错误：

 Fatal error: Call to undefined method DB_CONNECT::prepare();

我的理解是无法创建PDO对象。你能指点一下吗？

Answer 1

学习OOP不是创建无意义类的原因不幸的是，你创建了一个。 PDO不需要在其上构建类。请保持原样。

所以，而不是

$db = new DB_CONNECT("localhost", "oopcms","viktor","viktor");

制作

$db = new PDO("localhost", "oopcms","viktor","viktor");
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

这将更加清晰和有用

Answer 2

感兴趣的是这个和 Pekka 和你的常识的解决方案，谢谢您提示：）

 class Select
{
       private $query;
       private $dbh;
       private $row;

public function __construct(){

    $this->dbh =  new DB_CONNECT("localhost", "oopcms","viktor","viktor");



}


public function select(){

    $this->query = $this->dbh->db->prepare("SELECT * FROM `test`");
    $this->query->execute();
    $this->row = $this->query->fetchAll(PDO::FETCH_ASSOC);
    return $this->row;
  }


}

  $sel = new Select();
   $s = $sel->select();
  var_dump($s);

带有PDO语句的数据库连接类

2 个答案: