Symfony - Sonata Abstract Admin和getDoctrine

时间:2017-08-18 14:52:07

标签: php symfony doctrine sonata

我想使用prePersist()钩子将一个获取的对象设置为一个可以持久化的对象。但我无法想象如何在Sonata Admin Bundle中使用学说。

这是我的代码:

    namespace ShareBundle\Admin;

use Sonata\AdminBundle\Admin\AbstractAdmin;
use Sonata\AdminBundle\Datagrid\ListMapper;
use Sonata\AdminBundle\Datagrid\DatagridMapper;
use Sonata\AdminBundle\Form\FormMapper;



class UserShareAdmin extends AbstractAdmin
{
    protected function configureFormFields(FormMapper $formMapper)
    {
        $formMapper->add('quantity', 'text')
            ->add('user', 'sonata_type_model_list');
    }

    protected function configureDatagridFilters(DatagridMapper $datagridMapper)
    {
        $datagridMapper->add('quantity');
    }

    protected function configureListFields(ListMapper $listMapper)
    {
        $listMapper->addIdentifier('quantity')->addIdentifier('user')->addIdentifier('date');
    }

    public function prePersist($object)
    {
        $shareManager = $this->getDoctrine()->getManager()->getRepository('ShareBundle:Share');
        $value = $shareManager->findOneBy(array(), array('date' => 'DESC'));
        $object->setShare($value);
    }
}

有没有人知道怎么做?

谢谢!

1 个答案:

答案 0 :(得分:1)

好的伙计们,所以我已经弄清楚了!

我必须在服务参数

中指定orm默认实体管理器
#app/config/services.yml

arguments: [~, ShareBundle\Entity\UserShare, ~, @doctrine.orm.default_entity_manager]

我必须在Admin类中扩展构造函数。

 public function __construct($code, $class, $baseControllerName, $em)
{
    parent::__construct($code, $class, $baseControllerName);
    $this->em = $em;
}

(感谢this answer