Php Symfony中需要不同的类型

我试图用符号和教义创造一种形式。 我使用Doctrine在mysql中创建了一个作业类和一个与其相关的表。它还创建了JobType和JobController以及路由工具

我可以访问列出作业的索引页面，但无法访问新条目页面

以下是用于创建表单的文件

JobController.php

   <?php

namespace AppBundle\Controller;

use Symfony\Component\HttpFoundation\Request;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Method;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use AppBundle\Entity\Job;
use AppBundle\Form\JobType;

/**
 * Job controller.
 *
 * @Route("/job")
 */
class JobController extends Controller
{
    /**
     * Lists all Job entities.
     *
     * @Route("/", name="job_index")
     * @Method("GET")
     */
    public function indexAction()
    {
        $em = $this->getDoctrine()->getManager();

        $jobs = $em->getRepository('AppBundle:Job')->findAll();

        return $this->render('job/index.html.twig', array(
            'jobs' => $jobs,
        ));
    }

    /**
     * Creates a new Job entity.
     *
     * @Route("/new", name="job_new")
     * @Method({"GET", "POST"})
     */
    public function newAction(Request $request)
    {
        $job = new Job();
        $jobType = new JobType();
        $form = $this->createForm($jobType, $job);
        $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->persist($job);
            $em->flush();

            return $this->redirectToRoute('job_show', array('id' => $job->getId()));
        }

        return $this->render('job/new.html.twig', array(
            'job' => $job,
            'form' => $form->createView(),
        ));
    }

    /**
     * Finds and displays a Job entity.
     *
     * @Route("/{id}", name="job_show")
     * @Method("GET")
     */
    public function showAction(Job $job)
    {
        $deleteForm = $this->createDeleteForm($job);

        return $this->render('job/show.html.twig', array(
            'job' => $job,
            'delete_form' => $deleteForm->createView(),
        ));
    }

    /**
     * Displays a form to edit an existing Job entity.
     *
     * @Route("/{id}/edit", name="job_edit")
     * @Method({"GET", "POST"})
     */
    public function editAction(Request $request, Job $job)
    {
        $deleteForm = $this->createDeleteForm($job);
        $editForm = $this->createForm(new JobType(), $job);
        $editForm->handleRequest($request);

        if ($editForm->isSubmitted() && $editForm->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->persist($job);
            $em->flush();

            return $this->redirectToRoute('job_edit', array('id' => $job->getId()));
        }

        return $this->render('job/edit.html.twig', array(
            'job' => $job,
            'edit_form' => $editForm->createView(),
            'delete_form' => $deleteForm->createView(),
        ));
    }

    /**
     * Deletes a Job entity.
     *
     * @Route("/{id}", name="job_delete")
     * @Method("DELETE")
     */
    public function deleteAction(Request $request, Job $job)
    {
        $form = $this->createDeleteForm($job);
        $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->remove($job);
            $em->flush();
        }

        return $this->redirectToRoute('job_index');
    }

    /**
     * Creates a form to delete a Job entity.
     *
     * @param Job $job The Job entity
     *
     * @return \Symfony\Component\Form\Form The form
     */
    private function createDeleteForm(Job $job)
    {
        return $this->createFormBuilder()
            ->setAction($this->generateUrl('job_delete', array('id' => $job->getId())))
            ->setMethod('DELETE')
            ->getForm()
        ;
    }
}

谢谢

编辑： app/config/services.yml的内容

parameters:
#    parameter_name: value

services:
#    service_name:
#        class: AppBundle\Directory\ClassName
#        arguments: ["@another_service_name", "plain_value", "%parameter_name%"]

这在Symfony 3中不再可能。在Symfony 3中，始终必须传递表单类型的完全限定类名：

$editForm = $this->createForm(JobType::class, $job);

另外，在表单类型中，传递的是类型名称，而不是类型类的FQCN

---

Symfony 3刚刚发布了它的第一个测试版，这意味着它是非常前沿的。此外，Symfony 3几乎没有教程（因为它是非常前沿的）。您正在阅读Symfony 2教程，因此我建议您安装Symfony 2而不是3。

您能显示作业类型的服务定义吗这是我在app/config/services.yml

中找到的内容#了解有关服务的更多信息，#处的参数和容器http://symfony.com/doc/current/book/service_container.html 参数：#参数名称：值服务：#服务名称：#类：AppBundle\Directory\ClassName#参数：[“@other_service_name”、“plain_value”、“参数名称%”]我应该看看其他地方吗？考虑更新问题本身（在它下面有一个编辑按钮）和服务定义。正如您所看到的，注释的格式不好。从代码中可以清楚地看出，您没有一个JobType服务，这很好。我看不出你的代码有任何明显的错误。您可以尝试注释除一条之外的所有add语句以简化操作。我只留下了一条add语句，并尝试替换“string”、“text”和一条空语句，结果返回了相同的错误。从错误判断，似乎应该发送一个字符串来代替newjobType:
$form=$this->createForm（newjobType（），$job）$editForm = $this->createForm(new JobType(), $job);

$editForm = $this->createForm(JobType::class, $job);