Doctrine orm 权利联结主义

Doctrine orm 权利联结主义,doctrine-orm,symfony,doctrine-query,symfony-3.1,Doctrine Orm,Symfony,Doctrine Query,Symfony 3.1,我想加入symfony。我试过这里和这里的描述 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,

我想加入symfony。我试过这里和这里的描述

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();
看起来是这样的

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 = ?)
但我想问一下

raports
r右加入请求中选择* r、
requestraport
=rr.id


如何在doctrine2中使用右连接?

您可以像这样使用左连接并反转SQL

SELECT * FROM request_raports rr LEFT JOIN raports r ON r.requestRaports = rr.id
或者你也可以创建自己的“正确加入”。如果你看一下条令中的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)
    ...
我还没有测试过,不知道这是否是最好的方法

$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)
    ...