的Zend 2：使用分頁程序與TableGateway

Question

我在Module.php這樣來配置我的模型：

'MyModule\Model\MyTable' => function($sm) { 
    $tableGateway = $sm->get('MyTableGateway'); 
    $table = new MyTable($tableGateway); 
    return $table; 
}, 
'MyTableGateway' => function($sm) { 
    $dbAdapter = $sm->get('Zend\Db\Adapter\Adapter'); 
    $resultSetPrototype = new ResultSet(); 
    $resultSetPrototype->setArrayObjectPrototype(
     new Ticket()); 
    return new TableGateway('someTable', $dbAdapter, null, 
     $resultSetPrototype); 
    }, 

現在我這樣做是在MyTable.php：

$resultSet = $this->tableGateway->select(
    function(Select $select) { 
     $select 
      ->order('id ASC') 
      ->where->equalTo('aaaa', 2343) 
      ->and 
      ->where->notEqualTo('bbbb', 'sdfsdf') 
      ; 
}); 

現在我想使用分頁，但看zend tutorial他們有你這樣做：

// create a new Select object for the table album 
$select = new Select('someTable'); 
// create a new result set based on the Album entity 
$resultSetPrototype = new ResultSet(); 
$resultSetPrototype->setArrayObjectPrototype(new Album()); 
// create a new pagination adapter object 
$paginatorAdapter = new DbSelect(
    // our configured select object 
    $select, 
    // the adapter to run it against 
    $this->tableGateway->getAdapter(), 
    // the result set to hydrate 
    $resultSetPrototype 
); 
$paginator = new Paginator($paginatorAdapter); 
return $paginator; 

我爲什麼必須這樣做$select = new Select('someTable');這是否違反DI原則，因爲我靜態輸入表名？在Module.php中配置表名並且仍然必須在實際模型中對其進行硬編碼似乎是正確的。我能否以某種方式從服務管理器中檢索表名？

Answer 1

這裏是我的代碼看起來像現在按酥的評論

$select = new Select($this->tableGateway->getTable()); 

$paginatorAdapter = new DbSelect(
    $select, 
    $this->tableGateway->getAdapter(), 
    $this->tableGateway->getResultSetPrototype() 
); 
$paginator = new Paginator($paginatorAdapter); 

return $paginator; 

奇怪他們沒有做這樣的教程。