php - 尝试从命名空间 symfony Controller 调用函数

Question

我尝试在 Controller 中使用函数TechParentInsert，但出现错误:

Attempted to call function "TechParentInsert" from namespace "TestyBundle\Controller".

500 Internal Server Error - UndefinedFunctionException

在 Controller 中我有:“使用 TestyBundle\Repository\TechParentRepository；”当我定义“TechParentInsert”时

我做错了什么？

----------------------------我的代码

Controller TestyController:

namespace TestyBundle\Controller;

use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template;
use Symfony\Component\HttpFoundation\Request;
use Doctrine\ORM\Query\AST\Functions\ConcatFunction;
use Doctrine\ORM\EntityRepository;


use TestyBundle\Entity\Sort;
use TestyBundle\Form\SortType;
use TestyBundle\Repository\SortRepository;
use TestyBundle\Repository\TechParentRepository;

/**
 * @Route("/testy")
 *
 */
class TestyController extends Controller
{
    /**
     * (other actions )
     *
     */


    /**
     * @Route(
     *     "/parent/{parent}", name="TEST_sort_new"
     * )
     *
     * @Template
     */
    public function TechParentNewAction( Request $request, int $parent = 0 )
    {
            $repo2  = $this->getDoctrine()->getRepository( 'TestyBundle:Sort' );
            $sortNew = $repo2->findOneBy( ['zz' => 0]);

            $sortNew->setParentString( $sortNew->getParentString( ) . (string) $sortNew->getId() );
            $sortNew->setZz( 1 );

            $newRecordID = $sortNew->getId();

            $op1 = TechParentInsert( $newRecordID );

            $em->persist( $sortNew );
            $em->flush();

            return $this->redirect( $this->generateUrl( 'TEST_sort' ) );
        }

        return  [ 'data1' => $sortNew ];
    }

}

存储库:TechParentRepository

<?php
namespace TestyBundle\Repository;
use TestyBundle\Entity\TechParent;

class TechParentRepository extends \Doctrine\ORM\EntityRepository
{
    public function TechParentInsert( $idsort)
    {
        $parent = new TechParent();

        $parent->setIdSorty( $idsort);
        $parent->setIdParent( $idsort);
        $parent->setIsOwn( TRUE);

        $em = $this->getDoctrine()->getManager();

        $em->persist($parent);
        $em->flush();

        return 1;
    }
}

我的实体:TechParent

<?php
namespace TestyBundle\Entity;

use Doctrine\ORM\Mapping as ORM;

/**
 * TechParent
 *
 * @ORM\Table(name="tech_parent")
 * @ORM\Entity(repositoryClass="TestyBundle\Repository\TechParentRepository")
 */
class TechParent
{
/**
 * @var int
 *
 * @ORM\Column(name="id", type="integer")
 * @ORM\Id
 * @ORM\GeneratedValue(strategy="AUTO")
 */
private $id;

/**
 * @var int
 *
 * @ORM\Column(name="idSorty", type="integer")
 */
private $idSorty;

/**
 * @var int
 *
 * @ORM\Column(name="idParent", type="integer")
 */
private $idParent;

/**
 * @var bool
 *
 * @ORM\Column(name="isOwn", type="boolean")
 */
private $isOwn;


/**
 * Get id
 *
 * @return int
 */
public function getId()
{
    return $this->id;
}

/**
 * Set idSorty
 *
 * @param integer $idSorty
 *
 * @return TechParent
 */
public function setIdSorty($idSorty)
{
    $this->idSorty = $idSorty;

    return $this;
}

/**
 * Get idSorty
 *
 * @return int
 */
public function getIdSorty()
{
    return $this->idSorty;
}

/**
 * Set idParent
 *
 * @param integer $idParent
 *
 * @return TechParent
 */
public function setIdParent($idParent)
{
    $this->idParent = $idParent;

    return $this;
}

/**
 * Get idParent
 *
 * @return int
 */
public function getIdParent()
{
    return $this->idParent;
}

/**
 * Set isOwn
 *
 * @param boolean $isOwn
 *
 * @return TechParent
 */
public function setIsOwn($isOwn)
{
    $this->isOwn = $isOwn;

    return $this;
}

/**
 * Get isOwn
 *
 * @return bool
 */
public function getIsOwn()
{
    return $this->isOwn;
}
}

Answer 1

当您从不同的类访问方法时，您必须使用 ClassName::method()，但在这种情况下，TechParentRepository 肯定具有依赖性。由于依赖注入(inject)，Symfony 中有一个最佳实践将存储库定义为服务，然后以这种方式使用它:

$techParentRepository = $this->get('tech_parent_repository ');
$techParentRepository->TechParentInsert();

小建议 - 好的做法是将类命名为大写，但方法命名为小写，因此 public function techParentInsert($idsort) 会更好。