Php Can'；使用Zend TableGateway'；s select（）

我尝试使用TableGateway的

select（）

方法显示MySql表中的所有内容：

<section class="user_manager_index">
<h2>Users</h2>
   <?php
   $tableGateway = $this->tableGateway;
   $rowset = $tableGateway->select();

   foreach ($rowset as $row)
   {
       echo $row->id . ' ' . $row->name . ' ' . $row->email . '<br>';
   }
   ?>
</section>

其他信息：我的用户类：

class User
{
    public $id;
    public $name;
    public $email;
    public $password;

    public function setPassword($clear_password)
    {
        $this->password = md5($clear_password);
    }
    public function exchangeArray($data)
    {
        $this->name = (isset($data['name'])) ?
        $data['name'] : null;
        $this->email = (isset($data['email'])) ?
        $data['email'] : null;
        if (isset($data['password']))
        {
            $this->setPassword($data['password']);
        }
    }
}

---

您的

User

class

exchangeArray

方法用于用db值填充属性，它当前不引用

id

public function exchangeArray($data)
{
    // populate the id 
    $this->id = isset($data['id']) ? $data['id'] : null;

    $this->name = (isset($data['name'])) ?
    $data['name'] : null;
    $this->email = (isset($data['email'])) ?
    $data['email'] : null;
    if (isset($data['password']))
    {
        $this->setPassword($data['password']);
    }
}

---

非常感谢，正如您在代码片段中所示，我已经更改了类用户，现在我视图中的$row->id返回id值。

public function exchangeArray($data)
{
    // populate the id 
    $this->id = isset($data['id']) ? $data['id'] : null;

    $this->name = (isset($data['name'])) ?
    $data['name'] : null;
    $this->email = (isset($data['email'])) ?
    $data['email'] : null;
    if (isset($data['password']))
    {
        $this->setPassword($data['password']);
    }
}