如何在 Doctrine 中使用 conditionType 'ON'？

Question

我对 Doctrine 关于条件 ON 有疑问。我读过 here：

Now in doctrine 2 I have never seen a situation where ON can be used. In fact using ON always ends up in an exception saying you should use WITH.

但是我需要重新定义连接条件。这是我的查询：

$qb = $this->createQueryBuilder('fu');
$qb
    ->addSelect('cra')
    ->innerJoin('fu.chatRoomAdmins', 'cra')
    ->where('cra.operator = :operatorId')
    ->setParameter('operatorId', $operatorId);
$foreignUsers = $qb->getQuery()->getResult();

这里是（重要部分）生成的 sql 查询（我将 table 的名称更改为更具可读性）：

SELECT *
FROM   fu
       INNER JOIN cra
               ON fu.id = cra.operator_id
WHERE  ( cra.operator_id = 'an_id');

但我需要将 ON 条件更改为：ON fu.id = cra.operator_id OR cra.operator_id IS NULL

如果我像这样使用 WITH 条件：

$qb
//..
    ->innerJoin('fu.chatRoomAdmins', 'cra', 'WITH', 'cra.operator IS NULL')
//..
$foreignUsers = $qb->getQuery()->getResult();

它改变了我的查询：

SELECT *
FROM   fu
       INNER JOIN cra
               ON fu.id = cra.operator_id 
                  AND ( cra.operator_id IS NULL ) 
WHERE  ( cra.operator_id = 'an_id' );

但我想要这个 ON 条件：ON fu.id = cra.operator_id OR ( cra.operator_id IS NULL )

这就是为什么我试图在我的 queryBuilder 中用 ON 替换 WITH 的原因：

$qb
//..
    ->innerJoin('fu.chatRoomAdmins', 'cra', 'ON', 'cra.operator IS NULL')
//..
$foreignUsers = $qb->getQuery()->getResult();

但现在我得到一个错误：Expected end of string, got 'ON'（我使用的是 symfony 4）

你知道有什么方法可以达到这个 ON 条件：ON fu.id = cra.operator_id OR ( cra.operator_id IS NULL ) 吗？

Answer 1

我找到了一种方法 leftJoin :

$qb = $this->createQueryBuilder('fu');
$qb
    ->addSelect('cra')
    ->leftJoin('fu.chatRoomAdmins', 'cra')
    ->where('cra.operator = :operatorId OR cra.operator IS NULL');
$foreignUsers = $qb->getQuery()->getResult();

给这个 SQL:

SELECT *
FROM   fu 
       LEFT JOIN cra
              ON fu.id = cra.operator_id 
WHERE  ( cra.operator_id = 'an_id' 
          OR cra.operator_id IS NULL );

Answer 2

它是这样工作的

$query->innerJoin('App\Domains\Order\Domain\Models\Order\B2bOrderExt', 'b2b', \Doctrine\ORM\Query\Expr\Join::WITH, 'o.id=b2b.orderId');

如何在 Doctrine 中使用 conditionType 'ON'？

How to use conditionType 'ON' with Doctrine?

doctrine

doctrine-query

doctrine-orm

symfony4