Doctrine orm 权利联结主义
我想加入symfony。我试过这里和这里的描述 它给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,
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', $requestId->getFormId())
->getQuery();
看起来是这样的
SELECT
r0_.id AS id_0,
r0_.adminComment AS adminComment_1,
r0_.addDate AS addDate_2,
r0_.submitDate AS submitDate_3,
r0_.statusId AS statusId_4,
r0_.userId AS userId_5,
r0_.requestId AS requestId_6,
r0_.requestRaports AS requestRaports_7
FROM
raports r0_
INNER JOIN request_raports r1_ ON r0_.requestRaports = r1_.id
AND (r1_.formId = ?)
但我想问一下
从raports
r右加入请求中选择*
r、 requestraport
=rr.id
如何在doctrine2中使用右连接?您可以像这样使用左连接并反转SQL
SELECT * FROM request_raports rr LEFT JOIN raports r ON r.requestRaports = rr.id
或者你也可以创建自己的“正确加入”。如果你看一下条令中的leftJoin定义,它就像:
leftJoin($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)
...