namespace griffin\UserBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity(repositoryClass="griffin\UserBundle\Entity\UserRepository")
* @ORM\Table(name="users")
*/
class User {
const STATUS_ACTIVE = 1;
const STATUS_INACTIVE = 0;
/**
* @ORM\Id
* @ORM\Column(name="id_users", type="smallint")
* @ORM\GeneratedValue(strategy="AUTO")
*/
protected $idUsers;
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和Repository类一样
namespace griffin\UserBundle\Entity;
use Doctrine\ORM\EntityRepository;
class UserRepository extends EntityRepository {
public function getAdmin()
{
return $this->getEntityManager()
->createQuery('select * from users where users_groups_id = 1')
->getResults;
}
}
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当我在控制器中调用它时
$results = $this->getDoctrine()
->getRepository('griffin\UserBundle\Entity\UserRepository')
->getAdmin();
var_dump($results);
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我收到了错误
Class "griffin\UserBundle\Entity\UserRepository" sub class of "Doctrine\ORM\EntityRepository" is not a valid entity or mapped super class.
Mic*_*ick 12
getRepository() 将实体类作为第一个参数:
$results = $this->getDoctrine()
->getRepository('griffin\UserBundle\Entity\User')
->getAdmin();
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注意:快速查看EntityManager类本身总是一个好主意.如果这样做,您将看到此方法签名getRepository():
/**
* Gets the repository for an entity class.
*
* @param string $entityName The name of the entity.
*
* @return EntityRepository The repository class.
*/
public function getRepository($entityName)
{
//...
}
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