我想将当前用户传递给AbstractType。我按照此页面的帮助:Access currently logged in user in EntityRepository
不幸的是 - 这不适合我。我正在使用Symfony 2.6。
Catchable Fatal Error: Argument 1 passed to Checkout\Bundle\ItemBundle\Form\ItemType::__construct() must be an instance of Symfony\Component\Security\Core\SecurityContext, none given, called in /vagrant/src/Checkout/Bundle/ItemBundle/Controller/ItemController.php on line 231 and defined
那是我的类型:
<?php
namespace Checkout\Bundle\ItemBundle\Form;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;
use Symfony\Component\Security\Core\SecurityContext;
use Doctrine\ORM\EntityRepository;
class ItemType extends AbstractType
{
protected $securityContext;
public function __construct(SecurityContext $securityContext)
{
$this->securityContext = $securityContext;
}
/**
* @param FormBuilderInterface $builder
* @param array $options
*/
public function buildForm(FormBuilderInterface $builder, array $options)
{
$currentUser = $this->securityContext->getToken()->getUser();
$builder (...)
服务:
<?xml version="1.0" ?>
<container xmlns="http://symfony.com/schema/dic/services"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://symfony.com/schema/dic/services http://symfony.com/schema/dic/services/services-1.0.xsd">
<services>
<service id="form.type.item" class="Checkout\Bundle\ItemBundle\Form\ItemType">
<argument type="service" id="security.context" />
<tag name="form.type" alias="item" />
</service>
</services>
</container>
我试图弄清楚,但我并不是真的理解错误信息。有人可以帮忙吗? : - )
答案 0 :(得分:1)
您需要将服务标记为表单类型,并且只能使用它的别名。根据您的错误消息,您可能正在实例化该对象。在231左右发布ItemController的内容