从表单生成器symfony 2.8获取用户

Question

我想以选择的形式检索当前公司的框架列表，了解当前用户是公司。

但我有以下错误：

Attempted to call an undefined method named "get" of class "Gestion\StageBundle\Form\StageType".
 Did you mean to call e.g. "getBlockPrefix", "getName" or "getParent"?

my classType

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $ent_id= $this->get('security.context')->getToken()->getUser();
    $builder->add('sujet')->add('dateDebut')->add('dateFin')
    ->add('encadreur', EntityType::class, array(

'class' => 'UserBundle:Encadreur',
'query_builder' => function (EntityRepository $er) {
    return $er->createQueryBuilder('u')
        ->where('u.entreprise=$ent_id')
        ->orderBy('u.username', 'ASC');
},


'choice_label' => 'username',


));
  }

Answer 1

您可以将TokenStorage添加到表单类型构造函数

private $tokenStorage;

public function __construct(TokenStorageInterface $tokenStorage)
{
    $this->tokenStorage = $tokenStorage;
}

并使您的表单输入服务，例如

app.form.something:
        class: Gestion\StageBundle\Form\StageType
        arguments: ['@security.token_storage']
        tags:
            - { name: form.type }

然后您可以使用表单类型

获取用户

$user = $this->tokenStorage->getToken()->getUser();