Model Zend Framework 2模型外键
我现在正在接近Zend Framework 2,并遵循了这一点。一切都很好,但现在我想,例如,添加一个流派的专辑 我在数据库中添加了一个category表,并在相册表中创建了一个外键Model Zend Framework 2模型外键,model,zend-framework2,tablegateway,Model,Zend Framework2,Tablegateway,我现在正在接近Zend Framework 2,并遵循了这一点。一切都很好,但现在我想,例如,添加一个流派的专辑 我在数据库中添加了一个category表,并在相册表中创建了一个外键category\u id,它引用category表中的category.id 我如何使用TableGateway在我的模型上反映这种情况?我希望这将对您有所帮助,您将了解如何完成任务: <?php In module/Album/src/Album/Model/AlbumTable.php =========
category\u id
,它引用category表中的category.id
我如何使用
TableGateway
在我的模型上反映这种情况?我希望这将对您有所帮助,您将了解如何完成任务:
<?php
In module/Album/src/Album/Model/AlbumTable.php
==============================================
namespace Album\Model;
use Zend\Db\TableGateway\TableGateway;
class AlbumTable
{
protected $tableGateway;
public function __construct(TableGateway $tableGateway)
{
$this->tableGateway = $tableGateway;
$this->dbSql = new Sql($this->tableGateway->getAdapter()); // add this line
}
public function fetchAll()
{
$select = $this->dbSql->select();
$select->from('album')
->columns(array('id', 'artist', 'title'))
->join(array('C' => 'category'),
'fk_category = id_category',
array('id_category', 'name'),
$select::JOIN_LEFT);
$resultSet = $this->tableGateway->selectWith($this->select);
return $resultSet;
}
public function getAlbum($id)
{
$id = (int) $id;
$select = $this->dbSql->select();
$where = new Where();
$select->from('album')
->columns(array('id_album', 'artist', 'title'))
->join(array('C' => 'category'),
'fk_category = id_category',
array('id_category', 'name'),
$select::JOIN_LEFT);
$where->equalTo('id' => $id);
$rowset = $this->tableGateway->selectWith($this->select->where($where));
$row = $rowset->current();
if (!$row) {
throw new \Exception("Could not find row $id");
}
return $row;
}
public function saveAlbum(Album $album)
{
$data = array(
'artist' => $album->artist,
'title' => $album->title,
'fk_category' => $album->category_id
);
$id = (int)$album->id;
if ($id == 0) {
$this->tableGateway->insert($data);
} else {
if ($this->getAlbum($id)) {
$this->tableGateway->update($data, array('id' => $id));
} else {
throw new \Exception('Form id does not exist');
}
}
}
public function deleteAlbum($id)
{
$this->tableGateway->delete(array('id' => $id));
}
}
谢谢@prava。这是一个切实可行的解决办法。但是补水剂呢?我阅读了在TableGateway和模型类之间使用的。你知道怎么把它们联系起来吗?是的,我知道。我会在几天后更新我的答案。谢谢,非常有帮助。抱歉,如果我一直在问,但是,如果我想将Category建模为一个类,该怎么办?抱歉,没有理解您的观点@Danielanzelmi,请简单描述一下:)我想将Category类注入相册,而不是添加两个字段(CategoryId和CategoryName)。类似于$album->setCategory(Category$Category);
In module/Album/src/Album/Model/Album.php
=========================================
namespace Album\Model;
class Album
{
public $albumId;
public $artist;
public $title;
public $categoryId;
public $categoryName;
public function exchangeArray($data)
{
$this->albumId = (isset($data['id_album'])) ? $data['id_album'] : null;
$this->artist = (isset($data['artist'])) ? $data['artist'] : null;
$this->title = (isset($data['title'])) ? $data['title'] : null;
$this->categoryId = (isset($data['id_category'])) ? $data['id_category'] : null;
$this->categoryName = (isset($data['name'])) ? $data['name'] : null;
}
public function getArrayCopy()
{
return get_object_vars($this);
}
}
In module/Album/src/Album/Factory/Model/AlbumTableFactory.php
=============================================================
namespace Album\Factory\Model;
use Zend\ServiceManager\FactoryInterface;
use Zend\ServiceManager\ServiceLocatorInterface;
use Zend\Db\TableGateway\TableGateway;
use Album\Model\AlbumTable;
use Album\Model\Album;
use Zend\Stdlib\Hydrator\ObjectProperty;
use Zend\Db\ResultSet\HydratingResultSet;
class AlbumTableFactory implements FactoryInterface
{
public function createService(ServiceLocatorInterface $serviceLocator)
{
$db = $serviceLocator->get('Zend\Db\Adapter\Adapter');
$resultSetPrototype = new HydratingResultSet();
$resultSetPrototype->setHydrator(new ObjectProperty());
$resultSetPrototype->setObjectPrototype(new Album());
$tableGateway = new TableGateway('album', $db, null, $resultSetPrototype);
$table = new AlbumTable($tableGateway);
return $table;
}
}