select子句中的Doctrine dql子查询

Question

我尝试制作这样的查询构建器：

<?php 
$left = $this->em->getRepository(Recommendation::class)
->createQueryBuilder('r')
->select('count(r) as cnt')
->leftJoin('r.procedure', 'd')
->where('d.label is not NULL')
->andWhere('d.id in di.id')
->groupBy('d.label')
->getDQL()

$dict = $this->em->getRepository(\model\Dictionary\Dictionaryitem::class)
->createQueryBuilder('di')
->select('di.id, di.label')
->addSelect('(' .$left. ') as tt')
->join('di.dictionary', 'dd')
->where('dd.id = \'procedures\'')
->getQuery()
->getResult()
;

我得到的是：

  

预期的Doctrine \ ORM \ Query \ Lexer :: T_FROM，得到'标签'

“标签”指向以下片段“（SELECT p.label label ，count（r）”

我做错了什么？我记得，我从dql做了这样的子选择，它曾经工作过。

修改

我稍微修改了一下查询构建器，以简化消息。

我希望得到的SQL看起来像这样：

SELECT  di.id,  di.label, 
    (SELECT count(r) as cnt FROM recommendation as r 
    LEFT JOIN dictionary_dictionaryitem as d on r.procedure_id = di.id 
    where d.label is not null 
    and d.id = di.id 
    group by d.label) as tt 
from dictionary_dictionaryitem as di join dictionary_dictionary as dd on di.dictionary_id = dd.id where dd.id = 'procedures'
;

编辑2

问题解决了。以下代码确实有效。不确定是什么帮助，但它给出了我想要的结果。

$left = $this->em->getRepository(Recommendation::class)
->createQueryBuilder('r1')
->select('count(r1)')
->leftJoin('r1.procedure', 'p1')
->where('p1.label is not NULL')
->andWhere('r1.status in (:status)')
->andWhere('p1.id = di.id')
->groupBy('p1.label')
->getDQL()
;

$dict = $this->em->getRepository(\model\Dictionary\Dictionaryitem::class)
->createQueryBuilder('di')
->select('di.label')
->addSelect('(' . $left . ') as tt')
->join('di.dictionary', 'd')
->where('d.id = \'procedures\'')
->setParameters([
    'status' =>array_keys(RecommendationStatus::getRecommendationHasLeftStatuses())
])
->getQuery()
->getResult()
;