QueryBuilder與Symfony中的關系

我在 Symfony 中撰寫了一個搜索欄，它與一個名為“Structure”的物體的查詢構建器一起作業，該物體與一個物體“Partenaire”（OneToMany）相關，它作業得很好，但問題是它顯示了我需要的所有結構僅顯示與 Partenaire 相關的結構。如果有人可以幫我解決這個問題，謝謝。

PartenaireController.php：

#[Route('/{id}', name: 'app_partenaire_show', methods: ['GET', 'POST'])]
public function show(Partenaire $partenaire, EntityManagerInterface $entityManager, PartenaireRepository $partenaireRepository,Request $request, PartenairePermissionRepository $partenairePermissionRepository, StructureRepository $structureRepository): Response
{
    $getEmail = $this->getUser()->getEmail();
    $partenaireId = $entityManager->getRepository(Partenaire::class)->findOneBy([
        'id' => $request->get('id')
    ]);

    $search2 = $structureRepository->findOneBySomeField2(
        $request->query->get('q')
    );
    
    return $this->render('partenaire/show.html.twig', [
        'partenaire' => $partenaire,
        'permission'=>$partenairePermissionRepository->findBy(
            ['partenaire' => $partenaireId],
        ),
      'structures'=>$search2, // show all the structures

      /*  'structures'=>$structureRepository->findBy( // show the structure linked to the partenaire but doesn't work with the search
            ['partenaire' => $partenaireId],
            [],
        ),*/
        'email'=>$getEmail,
    ]);
}

結構存盤庫.php：

public function findOneBySomeField2(string $search2 = null): array
{
    $queryBuilder =  $this->createQueryBuilder('q')
        ->orderBy('q.id' , 'ASC');

    if ($search2) {
        $queryBuilder->andWhere('q.Adresse LIKE :search')
            ->setParameter('search', '%'.$search2.'%');
    }

    return $queryBuilder->getQuery()
        ->getResult()
        ;
}

uj5u.com熱心網友回復：

由于它是一對多關系，您應該加入正確的表并在結構查詢中添加 where 陳述句，例如：

public function findByPartenaireWithFilter(Partenaire $partenaire, string $search2 = null): array
{
    $queryBuilder =  $this->createQueryBuilder('q')
        ->leftJoin('q.partenaire','p')
        ->where('p.partenaire.id = :partenaireId')
        ->setParameter('partenaireId', $partenaire->getId())
        ->orderBy('q.id' , 'ASC');

    if (null !== $search2) {
        $queryBuilder->andWhere('q.Adresse LIKE :search')
            ->setParameter('search', '%'.$search2.'%');
    }

    return $queryBuilder
        ->getQuery()
        ->getResult();
}

uj5u.com熱心網友回復：

您需要Partenaire通過添加條件將結果限制在所需的范圍內。您可以通過簡單地將另一個引數傳遞給存盤庫函式來實作這一點。

我已經洗掉了控制器中的注釋代碼和 Partenaire 物體的加載，因為它應該已經加載，這要歸功于param converters的魔力。

#[Route('/{id}', name: 'app_partenaire_show', methods: ['GET', 'POST'])]
public function show(Partenaire $partenaire, EntityManagerInterface $entityManager, Request $request, PartenairePermissionRepository $partenairePermissionRepository, StructureRepository $structureRepository): Response
{
    $getEmail = $this->getUser()->getEmail();
    // The Partenaire should already be loaded

    $result = $structureRepository->findByPartenaireWithFilter(
         $partenaire,
         $request->query->get('q')
    );
    
    return $this->render('partenaire/show.html.twig', [
        'partenaire' => $partenaire,
        'permission' => $partenairePermissionRepository->findBy(
            ['partenaire' => $partenaireId],
        ),
        'structures' => $search2, // show all the structures
        'email' => $getEmail,
    ]);
}

至于存盤庫，它非常簡單：傳遞加載的partenaire 并添加一個where子句。

public function findByPartenaireWithFilter(Partenaire $partenaire, string $search2 = null): array
{
    $queryBuilder =  $this->createQueryBuilder('q')
        ->where('q.partenaire = :partenaire')
        ->setParameter('partenaire', $partenaire)
        ->orderBy('q.id' , 'ASC');

    if (null !== $search2) {
        $queryBuilder->andWhere('q.Adresse LIKE :search')
            ->setParameter('search', '%'.$search2.'%');
    }

    return $queryBuilder
        ->getQuery()
        ->getResult();
}

至于permissions控制器中的密鑰，我認為您要嘗試做什么，我建議您調查安全選民。

標籤：php交响乐关系查询生成器

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