这是我的“包装”类,比PDO(它更大,这是现在的重要部分)
class DB
{
protected $pdo;
protected $dsn;
protected $username;
protected $password;
protected $driverOptions;
protected $query;
public function __construct($dsn, $username = '', $password = '', array $driverOptions = [])
{
$this->dsn = $dsn;
$this->username = $username;
$this->password = $password;
$this->driver_options = $driverOptions;
}
public function connect()
{
$this->pdo = new PDO($this->dsn, $this->username, $this->password, (array)$this->driver_options);
$this->pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
return $this->pdo;
}
public function myQuery($sql)
{
$this->query = $this->pdo->prepare($sql);
$this->query->execute();
return $this->query;
}
public function all()
{
$all = $this->query->fetchAll();
$this->query->closeCursor();
return $all;
}
}
这完美地起作用(如果稍微改变,特别是通过添加类型提示),如下:
$class = new myPDO('mysql:host=localhost;dbname=***', 'login', 'pass');
$prepare = $class->connect()->prepare('SELECT * FROM test');
$prepare->execute();
$result = $prepare->fetch();
但我想用这种方式使用它:
$pdo = new DB('mysql:host=localhost;dbname=***', 'login', 'pass');
$result = $pdo->connect()->myQuery('SELECT * FROM test')->all();
我的IDE中出现此错误:
在PDO课程中找不到方法'myQuery'
答案 0 :(得分:2)
你的类充当了对PDO的抽象,但只要你写这个就泄漏了实际的PDO对象:
return $this->pdo;
这个类的消费者永远不需要知道该对象存在,所以你不应该给它们。
要获得链接效果,您所要做的就是让调用者返回他们已有的对象,准备再拨打电话。换句话说:
return $this;