Question

我正在建立一个网站，上面显示一些带有菜单的项目。在此菜单中，您可以选择一种语言来过滤项目。我的问题是查询以获取特定语言的项目。我有2个实体，即Project和Language：

项目：

<?php

namespace App\Entity;

use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\Common\Collections\Collection;
use Doctrine\ORM\Mapping as ORM;

/**
 * @ORM\Entity(repositoryClass="App\Repository\ProjectRepository")
 */
class Project
{
    /**
     * @ORM\Id()
     * @ORM\GeneratedValue()
     * @ORM\Column(type="integer")
     */
    private $id;

    /**
     * @ORM\ManyToMany(targetEntity="App\Entity\Language",     inversedBy="projects")
     */
    private $languages;

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


    /**
     * @return Collection|Language[]
     */
    public function getLanguages(): Collection
    {
        return $this->languages;
    }

    public function addLanguage(Language $language): self
    {
        if (!$this->languages->contains($language)) {
            $this->languages[] = $language;
        }

        return $this;
    }

    public function removeLanguage(Language $language): self
    {
        if ($this->languages->contains($language)) {
            $this->languages->removeElement($language);
        }

        return $this;
    }
}

语言：

<?php

namespace App\Entity;

use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\Common\Collections\Collection;
use Doctrine\ORM\Mapping as ORM;

/**
 * @ORM\Entity(repositoryClass="App\Repository\LanguageRepository")
 */
class Language
{
    /**
     * @ORM\Id()
     * @ORM\GeneratedValue()
     * @ORM\Column(type="integer")
     */
    private $id;

    /**
     * @ORM\ManyToMany(targetEntity="App\Entity\Project", mappedBy="languages")
     */
    private $projects;


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

    /**
     * @return Collection|Project[]
     */
    public function getProjects(): Collection
    {
        return $this->projects;
    }

    public function addProject(Project $project): self
    {
        if (!$this->projects->contains($project)) {
            $this->projects[] = $project;
            $project->addLanguage($this);
        }

        return $this;
    }

    public function removeProject(Project $project): self
    {
        if ($this->projects->contains($project)) {
            $this->projects->removeElement($project);
            $projet->removeLangage($this);
        }

        return $this;
    }
}

在我的数据库中，它们通过project_language相关联，并且在其中具有ID

这是我目前无法使用的查询：

$init = 'language_C_1';
$array = explode('_', $init);

$repo = $this->getDoctrine()->getRepository(Project::class);

$init = $array[2];
$query = $repo->createQueryBuilder('p')
    ->innerJoin('project_language', 'pl', Join::WITH, 'p.id = pl.project_id')
    ->innerJoin('language', 'l', Join::WITH, 'pl.language_id = l.id')
    ->where('l.id = :init')
    ->setParameter('init', $init)
    ->getQuery();

$arrayProjets = $query->getResult();

谢谢您的回答！

Answer 1

Doctrine允许您对数据库进行抽象。您不应该考虑数据库表，而应该考虑实体：您的项目实体具有$ languages属性，因此您可以像这样在您的queryBuilder中使用它：

$repo
->createQueryBuilder('p')
// add this to also load the languages entities
->addSelect('l')
->innerJoin('p.languages', 'l')
->where('l.id = :languageId')
->setParameter('languageId', $languageId)
->getQuery();

我对Product实体的属性$ languages进行了连接：连接表（此处为project_language）是完全透明的！

Symfony QueryBuilder

1 个答案: