Symfony2在控制器中找不到我的doctrine2实体类-如何解决？

Question

我试图在我的Symfony2控制器中运行一个简单的SQL语句（类似select * from table），但是它不起作用。 Symfony不知何故找不到课程。

一些信息：

1. 我试图在FROM子句中提供完整的名称空间+类名，而仅提供类名
2. 我已经尝试过DQL和QueryBuilder（请参见下面的代码。option1和option2）
3. AppKernel正在加载我的DoctrineBundle。 当我使用作曲家创建项目时已经存在
4. 我在settings.yml中尝试过auto_mapping是非
5. 我的代码段如下

错误信息：

[Semantical Error] line 0, col 14 near 'Job j ORDER BY': Error: Class 'Job' is not defined.
500 Internal Server Error - QueryException
1 linked Exception:

    QueryException »

[2/2] QueryException: [Semantical Error] line 0, col 14 near 'Job j ORDER BY': Error: Class 'Job' is not defined.  +
[1/2] QueryException: SELECT u FROM Job j ORDER BY j.name ASC  + 

settings.yml

doctrine:
    dbal:
        driver:   "%database_driver%"
        host:     "%database_host%"
        port:     "%database_port%"
        dbname:   "%database_name%"
        user:     "%database_user%"
        password: "%database_password%"
        charset:  UTF8
        # if using pdo_sqlite as your database driver, add the path in parameters.yml
        # e.g. database_path: "%kernel.root_dir%/data/data.db3"
        # path:     "%database_path%"

    orm:
        auto_generate_proxy_classes: "%kernel.debug%"
        auto_mapping: true
        #auto_mapping: false
        #mappings:
        #    MyAppMyBundle:
        #        type: annotation
        #        dir: Entity/

我的控制器

<?php

// src/MyApp/MyBundle/Controller/JobsController.php

namespace MyApp\MyBundle\Controller;

use MyApp\MyBundle\Entity\Job;

use Symfony\Bundle\FrameworkBundle\Controller\Controller;

class JobsController extends Controller {
    public function listAction() {
        $em = $this->getDoctrine()->getEntityManager();
        //$qb = $em->createQueryBuilder();

        //option1
        //$qb   ->select("j")
        //  ->from("Job", "j")
        //  ->orderBy("j.name", "ASC");*/
        //return $this->render('MyBundle:Jobs:list.html.twig', array('jobs' => $qb->getQuery()->getResult()));

        //option2
        $qb = $em->createQuery("SELECT u FROM Job j ORDER BY j.name ASC");
        return $this->render('MyBundle:Jobs:list.html.twig', array('jobs' => $qb->getResult()));
    }
}

我的实体班

<?php

// src/MyApp/MyBundle/Entity/Job.php

namespace MyApp\MyBundle\Entity;

use Doctrine\ORM\Mapping;

/**
 * @Mapping\Entity
 * @Mapping\Table(name="jobs")
 */
class Job {
    /**
     * @Mapping\Column(name="job_id", type="integer")
     * @Mapping\Id
     * @Mapping\GeneratedValue(strategy="AUTO")
     */
    protected $jobId;

    /**
     * @Mapping\Column(name="name", type="text")
     */
    protected $name;

    /**
     * @Mapping\Column(name="job_desc", type="text")
     */
    protected $description;

    /**
     * @Mapping\Column(name="personal_req", type="text")
     */
    protected $requirements;

    /**
     * Get jobid
     *
     * @return integer 
     */
    public function getJobId() {
        return $this->applicationId;
    }

    /**
     * Set name
     *
     * @param \text $name
     * @return Job
     */
    public function setName($name) {
        $this->name = $name;

        return $this;
    }
    /**
     * Get name
     *
     * @return text 
     */
    public function getName() {
        return $this->name;
    }

    /**
     * Set description
     *
     * @param \text $description
     * @return Job
     */
    public function setDescription($description) {
        $this->description = $description;

        return $this;
    }
    /**
     * Get description
     *
     * @return text 
     */
    public function getDescription() {
        return $this->description;
    }

    /**
     * Set requirements
     *
     * @param \text $requirements
     * @return Job
     */
    public function setRequirements($requirements) {
        $this->requirements = $requirements;

        return $this;
    }
    /**
     * Get requirements
     *
     * @return text 
     */
    public function getRequirements() {
        return $this->requirements;
    }

}

Answer 1

1. 如果直接通过entitymanager或仅通过MyAppMyBundle:Job进行查询，则使用完整的名称空间
2. 确保您的捆绑软件存在于AppKernel中
3. 倾向于使用$em->getRepository('MyAppMyBundle:Job')->createQueryBuilder('j')或$em->getRepository('MyAppMyBundle:Job')->findBy(array(),array('name' => 'ASC')
4. 使用php app/console doctrine:mapping:info和php应用程序/控制台学说：schema：validate验证模型

symfony中的异常总是很完美，因此请集中精力处理您的异常所说的内容

Answer 2

验证实体类的名称空间。 因为在写入错误的名称空间时没有生成错误，但是没有找到实体