doctrine-orm - 右连接原则

Question

我正在尝试正确加入 symfony。我按照此处所述Doctrine 2 - Outer join query和此处Symfony - Using Outer Joins with Doctrine ORM 进行了尝试。

$query = $em->getRepository('AppBundle:raports')->createQueryBuilder('r')
            ->select('r')
            ->leftJoin('r.requestRaports rr WITH rr.formId = :formId', false)
            ->setParameter('formId', $requestId->getFormId())
            ->getQuery();

它给

 SELECT 
  r0_.id AS id_0, 
  r0_.adminComment AS adminComment_1, 
  r0_.addDate AS addDate_2, 
  r0_.submitDate AS submitDate_3, 
  r0_.statusId AS statusId_4, 
  r0_.userId AS userId_5, 
  r0_.requestId AS requestId_6, 
  r0_.requestRaports AS requestRaports_7 
FROM 
  raports r0_ 
  LEFT JOIN request_raports r1_ ON r0_.requestRaports = r1_.id 
  AND (r1_.formId = ?)

当我尝试

$query = $em->getRepository('AppBundle:raports')->createQueryBuilder('r')
            ->select('r')
            ->join('r.requestRaports rr WITH rr.formId = :formId', false)
            ->setParameter('formId', $requestId->getFormId())
            ->getQuery();

看起来像这样

SELECT 
  r0_.id AS id_0, 
  r0_.adminComment AS adminComment_1, 
  r0_.addDate AS addDate_2, 
  r0_.submitDate AS submitDate_3, 
  r0_.statusId AS statusId_4, 
  r0_.userId AS userId_5, 
  r0_.requestId AS requestId_6, 
  r0_.requestRaports AS requestRaports_7 
FROM 
  raports r0_ 
  INNER JOIN request_raports r1_ ON r0_.requestRaports = r1_.id 
  AND (r1_.formId = ?)

但我想要像这样的查询

SELECT * FROM raportsr RIGHT JOIN request_raports rr ON r。requestRaports= rr.id

如何使正确的加入在教义2中起作用？

Answer 1

您可以像这样使用 LEFT JOIN 并反转您的 SQL

SELECT * FROM request_raports rr LEFT JOIN raports r ON r.requestRaports = rr.id

或者您可以创建自己的“正确连接”。如果你看一下学说中的 leftJoin 定义，它就像：

leftJoin($join, $alias, $conditionType = null, $condition = null, $indexBy = null)
{
    $parentAlias = substr($join, 0, strpos($join, '.'));

    $rootAlias = $this->findRootAlias($alias, $parentAlias);

    $join = new Expr\Join(
        Expr\Join::LEFT_JOIN, $join, $alias, $conditionType, $condition, $indexBy
    );

    return $this->add('join', array($rootAlias => $join), true);
}

所以，它可能看起来像这样：

$qb = $this->createQueryBuilder('b');

$rightJoin = new Expr\Join('RIGHT', 'r.requestRaports', 'rr', Expr\Join::WITH, 'rr.formId = :formId');
$qb
    ->select('r')
    ->add('join', ['r' => $rightJoin], true)
    ...

我没有测试过这个，不知道它是否是最好的方法......