Question

我有三个实体：Project，RequestMatcher和Response。以下是它们的相关性：

$project->getRequestMatchers(); // Project has many RequestMatchers

$requestMatcher->getActiveResponse(); // RequestMatcher has one active Response

如何编写DQL以获取Response的所有有效Project？

    $query = $this->getEntityManager()->createQuery(
        "SELECT resp FROM $responseClass resp
               JOIN $requestMatcherClass req WITH req.project = :project
               WHERE resp.id IN(req.active_response_id)"
    );

    $query->setParameter('project', $project);

我收到以下异常：

[Doctrine\ORM\Query\QueryException]                                 
  [Syntax Error] line 0, col 230: Error: Expected Literal, got 'req'

我也尝试了同样的结果：

    $query = $this->getEntityManager()->createQuery(
        "SELECT resp FROM $responseClass resp
               LEFT JOIN resp.requestMatcher req WITH req.project = :project
               WHERE resp IN(req.activeResponse)"
    );

这是将给我id的

的sql

select active_response_id from (
  select active_response_id from request_matchers where project_id = 1
) t

Answer 1

我不相信这是最好的方法，但我用以下方法解决了这个问题：

    $query = "SELECT resp FROM $responseClass resp
              WHERE resp.id IN (
                  SELECT IDENTITY(req.activeResponse) FROM $requestMatcherClass req
                  WHERE req.project = :project
              )";

Answer 2

您也可以尝试使用QueryBuilder对象，例如：

$this->getDoctrine()->getRepository('YourBundle:Project')->createQueryBuilder('p')
    ->join('p.requestMatchers', 'rm', \Doctrine\ORM\Query\Expr\Join::WITH, 'rm.project = :project')
    ->join('rm.activeReponse', 'ar')
    ->addSelect('rm.*')
    ->addSelect('ar.*')
    ->setParameter('project', $project)
    ->getQuery()
    ->execute();

通过加入与doctrine DQL的关系来选择对象

2 个答案: