在appDevDebugProjectContainer.php中找不到Symfony 2错误类 - Symfony 2 error class not found in appDevDebugProjectContainer.php

Symfony 2 error class not found in appDevDebugProjectContainer.php

我想添加具有安全上下文的表单服务如下：

namespace Infogold'AccountBundle'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 KonsultantType extends AbstractType {
     protected $user;
    public function __construct(SecurityContext $user) {

        $this->user = $user;
    }
 public function buildForm(FormBuilderInterface $builder, array $options) {
    -
    -
    -
    -
    -
    -
  public function getName() {
        return 'konsultanci_form';
    }

在服务中：

parameters:
#    infogold_account.example.class: Infogold'AccountBundle'Example
services:
   form.type.konsultanci_form:
       class:  Infogold'AccountBundle'Form'KonsulantType
       arguments: ['@security.context'] 
       tags:
            - { name: form.type, alias: konsultanci_form }

我收到此错误：

FatalErrorException： Error： Class 'Infogold''AccountBundle''Form''KonsulantType' 在 D：''xampp''htdocs''symfony4''app''cache''dev''appDevDebugProjectContainer.php 行 718 中找不到

718线是：

protected function getForm_Type_KonsultanciFormService()
    {
        return $this->services['form.type.konsultanci_form'] = new 'Infogold'AccountBundle'Form'KonsulantType($this->get('security.context'));
    }

我哪里犯了错误？

正如我所说，这是类名上的错误。它是KonsultantType而不是KonsulantType

尝试：

parameters:
#    infogold_account.example.class: Infogold'AccountBundle'Example
services:
   form.type.konsultanci_form:
       class:  Infogold'AccountBundle'Form'KonsultantType
       arguments: ['@security.context'] 
       tags:
            - { name: form.type, alias: konsultanci_form }