态
我试图在Symfony中正确加入。我尝试了此处描述的学说2-外部加入查询和Symfony-使用Outs of Outs与学说ORM。
$query = $em->getRepository('AppBundle:raports')->createQueryBuilder('r')
->select('r')
->leftJoin('r.requestRaports rr WITH rr.formId = :formId', false)
->setParameter('formId', $requestId->getFormId())
->getQuery();
它给出
SELECT
r0_.id AS id_0,
r0_.adminComment AS adminComment_1,
r0_.addDate AS addDate_2,
r0_.submitDate AS submitDate_3,
r0_.statusId AS statusId_4,
r0_.userId AS userId_5,
r0_.requestId AS requestId_6,
r0_.requestRaports AS requestRaports_7
FROM
raports r0_
LEFT JOIN request_raports r1_ ON r0_.requestRaports = r1_.id
AND (r1_.formId = ?)
当我尝试
时$query = $em->getRepository('AppBundle:raports')->createQueryBuilder('r')
->select('r')
->join('r.requestRaports rr WITH rr.formId = :formId', false)
->setParameter('formId', $requestId->getFormId())
->getQuery();
看起来像
SELECT
r0_.id AS id_0,
r0_.adminComment AS adminComment_1,
r0_.addDate AS addDate_2,
r0_.submitDate AS submitDate_3,
r0_.statusId AS statusId_4,
r0_.userId AS userId_5,
r0_.requestId AS requestId_6,
r0_.requestRaports AS requestRaports_7
FROM
raports r0_
INNER JOIN request_raports r1_ ON r0_.requestRaports = r1_.id
AND (r1_.formId = ?)
,但我想要像
这样的查询从
raports
r选择 *右加入request_raports rr on on r。requestRaports
= rr.id
如何使正确加入Doctrine2?
SELECT * FROM request_raports rr LEFT JOIN raports r ON r.requestRaports = rr.id
或者您可以创建自己的"正确加入"。如果您在学说中查看左键定义,就像:
leftJoin($join, $alias, $conditionType = null, $condition = null, $indexBy = null)
{
$parentAlias = substr($join, 0, strpos($join, '.'));
$rootAlias = $this->findRootAlias($alias, $parentAlias);
$join = new ExprJoin(
ExprJoin::LEFT_JOIN, $join, $alias, $conditionType, $condition, $indexBy
);
return $this->add('join', array($rootAlias => $join), true);
}
所以,看起来像这样:
$qb = $this->createQueryBuilder('b');
$rightJoin = new ExprJoin('RIGHT', 'r.requestRaports', 'rr', ExprJoin::WITH, 'rr.formId = :formId');
$qb
->select('r')
->add('join', ['r' => $rightJoin], true)
...
我尚未测试过,不知道它是最好的方法...