php - 在我自己的类中使用 mysqli_result::fetch_object 不起作用

Question

我将尝试使用 fetch_object('Classname') 来测试这个方法。

Inserent.php 中的类 Inserent:

namespace classes\model;

 class Inserent{
    private $nummer;
    private $nickname;
    private $email;

    public function __construct ($nummer, $nickname, $email){
        $this->nummer = $nummer;
        $this->nickname = $nickname;
        $this->email = $email;
    }

    public function getNummer(){
        return $this->nummer;
    }

    public function setNummer($nummer){
        $this->nummer = $nummer;
    }
 ...
 }

使用在 Mapper-Class InserentDAO.php 中:

namespace classes\mapper;
use classes\model\Inserent;

class InserentDAO{
    private $dbConnect;

    public function __construct (){
        $this->dbConnect = MySQLDatabase::getInstance();
    }

    public function readAll(){
        $sql = "SELECT inserentennummer, nickname, email FROM inserent";
        $inserent = null;
        $inserentList = array();

        if ($result = $this->dbConnect->query($sql)) {

            while ($inserent = $result->fetch_object('classes\model\Inserent')) {
                $inserentList[] = $inserent;
            }
            $result->close();
        }
        return $inserentList;
    }
...
}

我收到错误并且我的对象是空的:

Warning: Missing argument 1 for classes\model\Inserent::__construct() in D:\xampp\htdocs\workspace\secondStep\classes\model\Inserent.php on line 10

Warning: Missing argument 2 for classes\model\Inserent::__construct() in D:\xampp\htdocs\workspace\secondStep\classes\model\Inserent.php on line 10

Warning: Missing argument 3 for classes\model\Inserent::__construct() in D:\xampp\htdocs\workspace\secondStep\classes\model\Inserent.php on line 10

Notice: Undefined variable: nummer in D:\xampp\htdocs\workspace\secondStep\classes\model\Inserent.php on line 11

Notice: Undefined variable: nickname in D:\xampp\htdocs\workspace\secondStep\classes\model\Inserent.php on line 12

Notice: Undefined variable: email in D:\xampp\htdocs\workspace\secondStep\classes\model\Inserent.php on line 13

但是如果我在 Mapper 中更改代码，它将起作用。

部分代码 InserentDAO.php 代替:

while ($obj = $result->fetch_object()) {
     $inserent = new Inserent($obj->inserentennummer, $obj->nickname, $obj->email);
     $inserentList[] = $inserent;
}

Answer 1

我相信fetch_object调用一个空的构造函数，然后根据它获取的结果列设置属性。 There is a comment on the manual confirming that.

所以这就是fetch_object("classes\model\Inserent")是在做:

$object = new classes\model\Inserent();
$object->inserentennummer = $result['inserentennumer'];
$object->nickname = $result['nickname'];
$object->email = $result['email'];
return $object;

这就是为什么它提示缺少参数和 undefined variable (它们是 private )。

显而易见的解决方案是重构您的类以提供一个空的构造函数和公共(public)属性:

namespace classes\model;

class Inserent{
    public $nummer;
    public $nickname;
    public $email;

    public function __construct ($nummer = 0, $nickname = "", $email = ""){
        //...

使属性公开的另一种方法是使用 __set 魔术方法，但您仍然必须提供零参数构造函数。