Symfony2。库法不返回额外的字段（外键）对于许多一对多

我有方法在我的仓库类，但它没有继承外键的回报：Symfony2。库法不返回额外的字段（外键）对于许多一对多

public function findAll() 
{ 
    $qb = $this->createQueryBuilder('d'); 
    $result = $qb 
     ->select('d', 'corporation', 'restaurant', 'deviceType', 'deviceAccess') 
     ->leftJoin('d.corporation', 'corporation') 
     ->leftJoin('d.restaurant', 'restaurant') 
     ->leftJoin('d.deviceType', 'deviceType') 
     ->leftJoin('d.accesses', 'deviceAccess') 
     ->getQuery() 
     ->getArrayResult(); 

    return $result; 
}

deviceAccess这里是许多一对多的关系并且它也包含外键，但上面的方法不会返回这个外键。

我YAML配置：

AppBundle\Entity\Device: 
    type: entity 
    table: device 
    repositoryClass: AppBundle\Repository\DeviceRepository 

    manyToOne: 
     corporation: 
      targetEntity: Corporation 
      joinColumn: 
       onDelete: CASCADE 

     restaurant: 
      targetEntity: Restaurant 
      joinColumn: 
       onDelete: CASCADE 

     deviceType: 
      targetEntity: DeviceType 
      joinColumn: 
       onDelete: CASCADE 

    manyToMany: 
     accesses: 
      targetEntity: DeviceAccess 

    id: 
     id: 
      type: integer 
      id: true 
      generator: 
       strategy: AUTO 
    fields: 
     description: 
      type: text 
      nullable: true

和deviceAccess：

AppBundle\Entity\DeviceAccess: 
    type: entity 
    table: null 
    repositoryClass: AppBundle\Repository\DeviceAccessRepository 
    id: 
     id: 
      type: integer 
      id: true 
      generator: 
       strategy: AUTO 

    manyToOne: 
     deviceAccessType: 
      targetEntity: DeviceAccessType 
      joinColumn: 
       onDelete: CASCADE 

    fields: 
     address: 
      type: string 
      length: 255 
     login: 
      type: string 
      length: 255 
      nullable: true 
     password: 
      type: string 
      length: 255 
      nullable: true

方法不返回deviceAccessType场。 JSON结果例子：

[ 
    { 
    "id":34, 
    "description":"description1", 
    "corporation": { 
     "id":4, 
     "name":"corporation1", 
     "code":"74259", 
     "isDeleted":false 
    }, 
    "restaurant": { 
     "id":5, 
     "name":"restaurant1", 
     "address":"", 
     "code":"1234", 
     "isDeleted":false, 
     "isBlacklist":false 
    }, 
    "deviceType": { 
     "id":1, 
     "name":"printer", 
     "description":"printer" 
    }, 
    "accesses":[ 
     { 
    "id":22, 
    "address":"127.0.0.1", 
    "login":"admin", 
    "password":"password" 
     }, 
     { 
    "id":23, 
    "address":"192.168.0.0", 
    "login":"user", 
    "password":"password" 
     } 
    ] 
    } 
]

如何解决这个问题，并迫使库方法返回所有FOREIGH钥匙？

非常感谢您的帮助！

来源

2016-03-08 Sergio Ivanuzzo

只需要添加另外加入和更新的select语句： - > leftJoin（“deviceAccess.deviceAccessType”，'deviceAccessType）你也可以用预先加载鬼混，但加入是最直接的方式。 – Cerad

@Cerad非常感谢，这工作！请发表您的评论作为答案，我会接受它！ –

Arf ....真遗憾。 – chalasr

下应该做的伎俩：

$result = $qb 
    ->select('d', 'corporation', 'restaurant', 'deviceType', 'deviceAccess', 'deviceAccessType') 
    // ... 
    ->leftJoin('d.accesses', 'deviceAccess') 
    ->leftJoin('deviceAccess.deviceAccessType', 'deviceAccessType') 
    ->getQuery() 
    ->getArrayResult(); 

return $result;

来源

2016-03-08 16:49:41 chalasr

Symfony2。库法不返回额外的字段（外键）对于许多一对多

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