我已经阅读了有关此问题的各种解决方案,但似乎都没有效果。我真的想知道这个问题的核心是什么。我将准确列出我所做的事情,因为它相对简单,我无法理解我所缺少的东西。
所以我创建了一个带有表人的简单数据库,我正在尝试使用bootstrap生成CRUD,我工作正常。我的问题是当我尝试使用jquery插件来处理自动完成时。接下来我添加一个存储库来处理我的查询,当我得到Symfony2 Undefined方法'findLikeFullnameArray'消息时。我试图只使用注释,所以如果我的过程中出现问题,请告诉我。
以下是我的命令: 创建捆绑包
使用crud创建实体
然后我创建了我的SearchController:
namespace Company\NameofBundle\Controller;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template;
use Company\NameofBundle\Form\JqueryType;
use Company\NameofBundle\Form\SearchType;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\HttpFoundation\Request;
use Company\NameofBundle\Entity\Person;
/**
* Search controller.
*
* @Route("/search")
*/
class SearchController extends Controller
{
/**
* @Route("/", name="search")
* @Template()
*/
public function indexAction()
{
$form = $this->createForm(new SearchType(), null, [
'action' => '',
'method' => 'POST'
]);
return array(
'form' => $form->createView(),
);
}
/**
* @Route("/person_search", name="person_search")
* @Template()
*
* @param Request $request
*
* @return array
*/
public function searchPersonAction(Request $request)
{
$q = $request->get('term');
$em = $this->getDoctrine()->getManager();
$results = $em->getRepository('CompanyNameofBundle:Person')->findLikeFullname($q);
return array('results' => $results);
}
/**
* @Route("/person_get", name="person_get")
*
* @param $id
*
* @return Response
*/
public function getPersonAction($id)
{
$em = $this->getDoctrine()->getManager();
$book = $em->getRepository('CompanyNameofBundle:Person')->find($id);
return new Response($book->getFullname());
}
/**
* @Route("/jquery", name="jquery")
* @Template()
*/
public function jqueryAction()
{
$form = $this->createForm(new JqueryType(), null, [
'action' => '',
'method' => 'POST'
]);
return array(
'form' => $form->createView(),
);
}
/**
* @Route("/jquery_search/{phrase}", name="jquery_search")
*
* @param string $phrase
*
* @return JsonResponse
*/
public function searchJqueryAction($phrase)
{
$em = $this->getDoctrine()->getManager();
$results = $em->getRepository('CompanyNameofBundle:Person')->findLikeFullnameArray($phrase);
return new JsonResponse($results);
}
}
人员实体:
<?php
namespace Company\NameofBundle\Person;
use Doctrine\ORM\Mapping as ORM;
/**
* Person
* @ORM\Table()
* @ORM\Entity(repositoryClass="Company\NameofBundle\Entity\Repository\PersonRepository")
*/
class Person
{
/**
* @var integer
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @var string
*
* @ORM\Column(name="fullname", type="string", length=255)
*/
private $fullname;
/**
* @var string
*
* @ORM\Column(name="email", type="string", length=255)
*/
private $email;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* Set fullname
*
* @param string $fullname
* @return Person
*/
public function setFullname($fullname)
{
$this->fullname = $fullname;
return $this;
}
/**
* Get fullname
*
* @return string
*/
public function getFullname()
{
return $this->fullname;
}
/**
* Set email
*
* @param string $email
* @return Person
*/
public function setEmail($email)
{
$this->email = $email;
return $this;
}
/**
* Get email
*
* @return string
*/
public function getEmail()
{
return $this->email;
}
}
最后是PersonRepository
<?php
namespace Company\NameofBundle\Entity\Repository;
use Doctrine\ORM\EntityRepository;
class PersonRepository extends EntityRepository
{
public function findLikeFullnameArray($fullname)
{
return $this->createQueryBuilder('person_repository')
->where('person_repository.fullname LIKE :name')
->setParameter('name', '%' . $fullname . '%')
->getQuery()
->getArrayResult();
}
}
以防这是我的app / config / routing.yml
company_nameofbundle:
resource: "@CompanyNameofBundle/Controller/"
type: annotation
prefix: /
提前致谢!
答案 0 :(得分:0)
第一次,您应该查看文档以获得这样的存储库:
http://symfony.com/doc/current/book/doctrine.html
AudioRecord
第二次,您必须检查已放置的// src/AppBundle/Entity/Product.php
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity(repositoryClass="AppBundle\Entity\ProductRepository")
*/
class Product
{
//...
}
格式
xml
您不能混合使用<entity name="Acme\StoreBundle\Entity\Product" repository-class="Acme\StoreBundle\Entity\ProductRepository">
,annotation
或xml
格式来塑造Symfony 2中的实体,删除所有不必要的文件。