Php mysqli及其继承的错误

我尝试用名为Database的继承来扩展mysqli类，如果操作正确，它可以在一行中轻松检索数据库信息。到目前为止还不可行，因为我得到了以下错误：

以下是定义的数据库类：

class Database extends mysqli{
  private $select, $create, $insert, $alter, $update, $delete, $drop;
  protected $mysql, $result, $table, $column, $where, $value, $limit, $order, $type;

  public function __construct($host, $user, $pass, $db){
    $this->mysql = new mysqli($host, $user, $pass, $db) 
           or die("Error connecting to the database {$db}");

  }

  public function __destruct(){
    $this->mysql->close();
  }

  protected function prepareQuery(){
    if ($prepare = $this->mysqli->prepare($this->query)) {
      trigger_error("Problem preparing query ($this->query) ".$this->mysqli->error, E_USER_ERROR);
    }
    return $prepare;
  }

  protected function reset(){
    unset($this->table);
    unset($this->column);
    unset($this->where);
    unset($this->value);
    unset($this->limit);
    unset($this->order);
  }

  public function select($column){    
    $this->select = implode(",", $column);
    return $this;
  }

  public function table($table){
    $this->table = $table;
    return $this;
  }

  public function where($where, $comparison = "=", $logic){
    $i = 0;
    foreach ($where as $col => $val){
      $wherestring .= (is_array($comparison)) ? " {$col} {$comparison[$i]} '{$val}'" : " WHERE {$col} {$comparison} '{$val}'";
      $wherestring .= ($i < (count($where)-1))?" {$logic[$i]}" :" ";
      $i++;
    }
    $this->where = $wherestring;
    return $this;
  }

  public function limit($limit){
    $this->limit = $limit;
    return $this;
  }

  public function order($order){
    $this->order = $order;
    return $this;
  }

  public function runquery($method){
    $query = "{$method} {$this->select} FROM {$this->table}";
    if(!empty($this->where)) $query .= " WHERE {$this->where}";
    if(!empty($this->limit)) $query .= " LIMIT {$this->limit}";
    if(!empty($this->order)) $query .= " ORDER BY {$this->order}";
    echo "The generated Query is: \n".$query;
    $this->result = parent::query($query);
    $result = parent::fetch_array($this->result);
    return $result;
  } 

}

它不起作用，我收到以下警告和错误消息： 代码：

我现在有点困惑，因为我似乎无法找到解决这个问题的方法。你们谁能帮忙吗？我非常感激

谢谢

fecth_数组不是mysqli的方法，而是mysqli_STMT类

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你们不应该这样设计，你们在做扩展mysqli的事情，但你们也在做$this->mysql=newmysqli…，在is-A和has-A之间，你们应该选择一个。我推荐has-A.

除了xdazz所说的，您还混合了“继承自”和“使用”模式：您有一个名为$this->mysql的mysqli实例，以及$this，继承自父：：mysqli。你应该选择其中一个，但不能同时使用两个。如果您希望您的类的用户能够用它做任何她可以用普通mysqli实例做的事情，包括自己调用所有超类方法，请使用继承。如果要限制用例，请使用mysqli类型实例变量并删除继承。然后，您可以有选择地代理内部$this->mysql的方法，以允许用户调用它们。

哦，我的答案刚刚被取代，xdazz现在也这么说…

include("inc/config.php");
include("classes/class_data.php");

$db = new Database($dbhost, $dbuser, $dbpass, $dbname);
$row = $db->select(array("password","email"))->table($prefix."users")->where(array("uid"=>1, "username"=>Admin"),array("=","="),array("AND"))->limit(2)->order("uid")->runquery("SELECT");

Warning: mysqli::query() [mysqli.query]: Couldn't fetch Database in classes/class_data.php on line 70
Fatal error: Call to undefined method mysqli::fetch_array() inclass_data.php on line 71