Question

我有非常基本的课程：

class Customer {

  protected $id;
  protected $customer;

  public function __construct($customer_id) {
    $this->id = $customer_id;
    return $this->set_customer();
  }

  protected function set_customer() {
    $query = mysql_query("SELECT * FROM customer WHERE id = '$this->id'");
    $this->customer = mysql_fetch_row($query);
    return $this->customer;
  }
}

$customer = new Customer($order->customer->id);
print_r($customer);

这不是我想要的，但我理解为什么...... $ customer返回对客户对象的引用......

我想要的是来自mysql_fetch_row（）的MySQL行数组 - 如何返回数组？

我错过了什么？

Answer 1

当然。如果构造一个新对象，则会返回对新对象的引用。你怎么能保留它的参考？

$customer = new Customer();
// if $customer was a data array, where did the object reference go?

new运算符将始终返回对新创建的对象的引用。你不能从构造函数中返回任何其他内容。

Answer 2

您可以利用ArrayObject（假设您使用的是PHP 5.3）来获得您想要的内容：

class Customer extends ArrayObject {

    protected $id;
    protected $customer;

    public function __construct($customer_id) {
        $this->id = $customer_id;
        $this->set_customer();
        parent::__construct($this->customer);
    }

    protected function set_customer() {
       $query = mysql_query("SELECT * FROM customer WHERE id = '$this->id'");
       $this->customer = mysql_fetch_row($query);
    }
}

然后您可以将$customer用作数组：

$customer = new Customer($customer_id);
echo $customer['name'];

Answer 3

听起来你想要一个静态方法（根据评论判断） - 最后看一下例子。

但是，我认为你错过了对象的'心态'。

类客户应该是一个对象 - 一个真正的客户。你已经到了一半，但是你正在拿一个阵列把它撞成一个空白物体。

使用set_customer来填充你的对象 - 就像这样（这是在浏览器中键入的顺便说一下，只是想到了这个想法）：

class Customer {

  public $id;
  public $name;
  public $address;

  public function __construct($customer_id) {
    $this->customer = $this->set_customer($id);
  }

  private function set_customer($id ) {
    $query = mysql_query("SELECT * FROM customer WHERE id = '$id'");
    $customer = mysql_fetch_assoc($query);
    foreach($customer as $field => $value) {
       $this->$field = $value;
    }
  }
}

$customer = new Customer();
print_r($customer); // now has $customer->id, $customer->name, $customer->address

以适合您的代码和评论：

class Customer {

  protected $id;
  public $customer;

  public function __construct($customer_id) {
    $this->id = $customer_id;
    $this->customer = self::get_customer($id);
  }

  public static function get_customer($id ) {
    $query = mysql_query("SELECT * FROM customer WHERE id = '$id'");
    return mysql_fetch_row($query);
  }
}

$customer = Customer::get_customer($id);

Answer 4

这个怎么样：

class Customer {

  protected $id;

  public function __construct($customer_id) {
    $this->id = $customer_id;
  }

  public function getAsArray() {
    $query = mysql_query("SELECT * FROM customer WHERE id = '$this->id'");
    return mysql_fetch_assoc($query);
  }
}

$customer = new Customer( $id );
$customer_array = $customer->getAsArray();
print_r( $customer_array );

那就说了几件事：

该类缺少很多（错误/异常处理）。
你可能想要查看php的许多ORM's中的一个。
您可能希望将mysql_ *替换为PDO并使用参数化查询。

Answer 5

以下是我认为更面向对象的风格：

class CustomerFind {

    public static function byIdNumber($customer_id) {
      $query = mysql_query("SELECT * FROM customer WHERE id = '$customer_id'");
      return mysql_fetch_assoc($query);
    }

}

$customer = CustomerFind::byIdNumber( 1 );
print_r( $customer );

Answer 6

将结果放在类中的公共变量中，然后在$ customer上执行print_r，不要在构造函数中返回。

目前它是受保护的变量，所以你不会用print_r

看到它

这样的事情有用吗？

class Customer {

  protected $id;
  public $customer;

  public function __construct($customer_id) {
    $this->id = $customer_id;
    $this->set_customer();
  }

  protected function set_customer() {
    $query = mysql_query("SELECT * FROM customer WHERE id = '$this->id'");
    $this->customer = mysql_fetch_row($query);
  }
}

$customer = new Customer($order->customer->id);
print_r($customer);

Basic类返回对象引用而不是Array

6 个答案: