从数据库中检索数据并将其作为实体实例传递

Retrieving data from database and passing it as instance of entity

我正在尝试将 id 从一个数据库 table 的查询传递到另一个 table 相关的数据库。
但是,由于它们是相关的,我必须将其作为相关实体 (class) 的实例传递。
这是我得到的错误:

Argument 1 passed to App\Entity\KeywordRating::setKeywordId() must be an instance of App\Entity\Keywords, integer given, called in C:\xampp\htdocs\ratingAPI\src\Service\KeywordRatingService.php on line 26

这是我的代码:

public function insertRating($params, ObjectManager $manager, KeywordsRepository $keyRep, ProvidersRepository $provRep)
{
    $keywordId = $keyRep->checkKeyword(strtolower($params['keyword']))->getId();
    $providerId = $provRep->checkProvider($params['provider'])->getId();

    $rating = new KeywordRating();
    $rating->setKeywordId(<Keywords instance> $keywordId);
    $rating->setProviderId(<Providers instance> $providerId);
    if($params['attr'] === 'rocks')
    {
        $rating->setRocksRating($params['rating']);
        $rating->setSucksRating(0);
    } else 
    {
        $rating->setRocksRating(0);
        $rating->setSucksRating($params['rating']);
    }
    $rating->setRocksVotes(10);
    $rating->setSucksVotes(8);
    $rating->setSumVotes(18);
    $rating->setScore(2.50);

    $manager->persist($rating);
    $manager->flush();
}

这是sql查询:

public function checkKeyword($keyword)
{
    return $this->createQueryBuilder('k')
        ->where('k.keyword_name = :val')
        ->setParameter('val', $keyword)
        ->getQuery()
        ->getOneOrNullResult();
}

checkProvider()是一样的,只是参数不同。

我该怎么做?

假设您有一个名为 Entities 的命名空间,其中包含所有实体 类,您可以尝试这样的操作:

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

    $queryBuilder = $entityManager->createQueryBuilder();
    $queryBuilder->select('keyword')
                ->from('App\Entity\Keywords', 'keyword')
                ->where($queryBuilder->expr()->eq('keyword.keyword_name', $queryBuilder->expr()->literal($keyword)));

    $result = $queryBuilder->getQuery()-> getOneOrNullResult();

    return $result;