doctrine-orm - Doctrine DQL 返回多种类型的实体

Question

我有三个实体:HandsetSubscription、Handset 和 Subscription。

HandsetSubscription 的 yaml 是:

App\SoBundle\Entity\HandsetSubscription:
type: entity
table: handset_subscription
manyToOne:
    handset:
        targetEntity: Handset
    subscription:
        targetEntity: Subscription                    
id:
    id:
        type: integer
        generator: { strategy: AUTO }
        options: { unsigned: true }
fields:
    amount:
        type: integer
        nullable: false
        options: { default: 0, unsigned: true  }
    discount:
        type: integer
        nullable: false
        options: { default: 0, unsigned: true  }

查询:

SELECT hs,s,h
      FROM  \App\SoBundle\Entity\HandsetSubscription hs                    
      JOIN  \App\SoBundle\Entity\Subscription        s with s.id    = hs.subscription 
                        AND s.mins  = 150 
                        AND s.mb    = 250 
                        AND s.sms   = 150
      JOIN  \App\SoBundle\Entity\Handset             h with h.id    = hs.handset ​

这些是检索到的条目的类名:

App\SoBundle\Entity\HandsetSubscription
Proxies\__CG__\App\SoBundle\Entity\Subscription
Proxies\__CG__\App\SoBundle\Entity\Handset
App\SoBundle\Entity\HandsetSubscription
Proxies\__CG__\App\SoBundle\Entity\Handset
App\SoBundle\Entity\HandsetSubscription
Proxies\__CG__\App\SoBundle\Entity\Handset
…

我希望只返回 HandsetSubscription 实体。为什么我也得到订阅和手机的代理？

通过在手机和订阅映射中添加 fetch eager 并从查询中的 SELECT 语句中删除手机和订阅，我只会获得 HandsetSubscription，但我想通过 fetch 连接来完成此操作，如手册中所述(http://doctrine-orm.readthedocs.org/en/latest/reference/dql-doctrine-query-language.html#joins ).

更新

引用上面发布的链接:

获取地址的连接:

<?php
$query = $em->createQuery("SELECT u, a FROM User u JOIN u.address a WHERE a.city = 'Berlin'");
$users = $query->getResult();

当 Doctrine 使用 fetch-join 组合查询时，它会在结果数组的根级别返回 FROM 子句中的类。在前面的示例中，返回了一个 User 实例数组，每个用户的地址都被提取并保存到 User#address 变量中。如果您访问该地址，Doctrine 不需要延迟加载与另一个查询的关联。

Answer 1

非常感谢来自#doctrine irc channel 的 veonik 解决了这个问题。

您应该加入协会，而不是加入实体的完全限定名称。所以查询变成:

SELECT hs,s,h
  FROM  \App\SoBundle\Entity\HandsetSubscription hs                    
  JOIN  hs.subscription s with s.id = hs.subscription 
                    AND s.mins  = 150 
                    AND s.mb    = 250 
                    AND s.sms   = 150
  JOIN  hs.handset h with h.id = hs.handset