Question

我希望以一对多的关系访问实体的伟大孩子。例如：部门实体（父母） - ＆gt;类别实体（儿童） - ＆gt;组实体（大孩子） - ＆gt;产品实体（曾孙子）。我想根据他们的部门获取产品。

部门与类别的关系：

/**
 *  @ORM\OneToMany(targetEntity="Category", mappedBy="department")
 */
protected $category;
public function __construct()
{
    $this->category = new ArrayCollection();
}

与部门和小组的类别关系：

 /**
 * @ORM\ManyToOne(targetEntity="Department", inversedBy="category")
 * @ORM\JoinColumn(name="department_id", referencedColumnName="id")
 */
protected $department;
/**
 *  @ORM\OneToMany(targetEntity="Group", mappedBy="category")
 */
protected $group;
public function __construct()
{
    $this->group = new ArrayCollection();
}

与类别和产品的群组关系：

/**
 * @ORM\ManyToOne(targetEntity="Category", inversedBy="group")
 * @ORM\JoinColumn(name="category_id", referencedColumnName="id")
 */
protected $category;

/**
 *  @ORM\OneToMany(targetEntity="Product", mappedBy="group")
 */
protected $product;
public function  construct()
{
    $this->product = new ArrayCollection();
}

/**
 * @ORM\ManyToMany(targetEntity="Brand", inversedBy="group")
 * @ORM\JoinTable(name="brand_groups")
 */
private $brand;

public function __construct()
{
    $this->brand = new ArrayCollection();
}

与群组的产品关系：

/**
 * @ORM\ManyToOne(targetEntity="Group", inversedBy="product")
 * @ORM\JoinColumn(name="groups_id", referencedColumnName="id")
 */
protected $group;

ProductRepository：

namespace AppBundle\Repository;

use Doctrine\ORM\EntityRepository;
use AppBundle\Entity\Product;

/**
 * ProductRepository
 */
class ProductRepository extends EntityRepository
{
    public function findDepartmentProduct($department)
    {
        return $this
            ->createQueryBuilder('p')
            ->select(array('p', 'g','c','d'))
            ->from('Product', 'p')
            ->join('p.Group', 'g')
            ->join('g.Category', 'c')
            ->join('c.Department', 'd')
            ->where('d = :department')
            ->setParameter('department', $department)
            ->getQuery();
    }
}

DefaultController：

/**
 * @Route("/department/{id}", name = "department")
 */
public function departmentAction(Request $request,$id)
{
    $Department = $this->getDoctrine()->getRepository('AppBundle:Department');
    $department = $Department->find($id);

    $Product = $this->getDoctrine()->getRepository('AppBundle:Product');
    $customersChoiceProducts = $Product->mostView('20');

    $query = $query = $Product->findDepartmentProduct($department);

    $paginator  = $this->get('knp_paginator');
    $pagination = $paginator->paginate(
        $query, 
        $request->query->getInt('page', 1),
        8/*limit per page*/
    );

    return $this->render('default/department.html.twig', array(
        'pagination' => $pagination,
        'department' => $department,
        'customersChoiceProducts' => $customersChoiceProducts,
    ));
}

然后我收到此错误：

[语义错误]第0行，第63行靠近'g，产品p'：错误：类AppBundle \ Entity \ Product没有名为groups的关联

请帮帮我。

Answer 1

我犯的错误是加入实体类名而不是关联。我的新代码是：

public function findDepartmentProduct($department)
{
    return $this
        ->createQueryBuilder('l')
        ->select('p')
        ->from('AppBundle:Product','p')
        ->join('p.group', 'g')
        ->join('g.category', 'c')
        ->join('c.department', 'd')
        ->where('d = :department')
        ->setParameter('department', $department)
        ->getQuery();
}

如果你运行它，你会收到以下错误：

无法统计选择两个FROM组件的查询，无法区分

您可以通过添加->getResult()

来解决此错误

public function findDepartmentProduct($department)
{
    return $this
        ->createQueryBuilder('l')
        ->select('p')
        ->from('AppBundle:Product','p')
        ->join('p.group', 'g')
        ->join('g.category', 'c')
        ->join('c.department', 'd')
        ->where('d = :department')
        ->setParameter('department', $department)
        ->getQuery()
        ->getResult();
}

语义错误行0，col 63靠近g，产品p：错误：类AppBundle \ Entity \ Product没有关联symfony2应用程序

1 个答案: