Symfony的实体加盟问题

Question

林遇到了左边的一个问题，同时加入执行这个查询：

$result = $this->createQueryBuilder('a') 
       ->select(['a','u']) 
       ->leftJoin(User::class, 'u', Join::WITH, 'a.user_id = u.id') 
       ->where('a.api_key = :api_key') 
       ->setParameter('api_key', $key) 
       ->getQuery() 
       ->getArrayResult(); 

以下是错误：

[Semantical Error] line 0, col 84 near 'user_id = u.id': Error: Class AppBundle\Entity\ApiKey has no field or association named user_id 

user_ID的存在为关联的名字，所以我不知道为什么我看到这个错误？

这里是我的实体：

user.php的

/** 
* User 
* 
* @ORM\Table("users") 
* @ORM\Entity 
* @UniqueEntity(
*  fields={"username"}, 
*  message="This username is already in use." 
*) 
* @UniqueEntity(
*  fields={"email"}, 
*  message="This email address is already in use." 
*) 
* @ORM\Entity(repositoryClass="AppBundle\Repository\UserRepository") 
*/ 
class User 
{ 
/** 
* @var integer 
* @ORM\Column(name="id", type="integer") 
* @ORM\Id 
* @ORM\GeneratedValue(strategy="AUTO") 
*/ 
protected $id; 

/** 
* @ORM\Column(name="username", type="string", length=225, unique=true) 
* @Assert\NotBlank(message="Please enter a username.") 
* @Assert\Length(
*  min = 2, 
*  max = 225, 
*  minMessage = "Your username must be at least {{ limit }} characters", 
*  maxMessage = "Your username cannot be longer than {{ limit }} characters" 
*) 
*/ 
protected $username; 

/** 
* @ORM\Column(name="email", type="string", length=225, unique=true) 
* @Assert\NotBlank(message="Please enter a email.") 
* @Assert\Email(message = "The email '{{ value }}' is not a valid email.") 
*/ 
protected $email; 

/** 
* @ORM\Column(name="salt", type="text") 
*/ 
protected $salt; 

/** 
* @ORM\Column(name="password", type="string", length=225) 
* @Assert\NotBlank(message="Please enter a password.") 
* @Assert\Length(
*  min = 7, 
*  minMessage = "Your password must be at least {{ limit }} characters long." 
*) 
*/ 
protected $password; 

/** 
* @Assert\NotBlank(message="Please enter a confirmation password.") 
*/ 
protected $confirmationPassword; 

/** 
* @ORM\Column(name="roles", type="json_array") 
*/ 
protected $roles; 

/** 
* @ORM\OneToOne(targetEntity="ApiKey", mappedBy="user", cascade={"all"}, fetch="LAZY") 
*/ 
protected $apiKey; 

/** 
* @ORM\OneToOne(targetEntity="Nation", mappedBy="user", cascade={"all"}, fetch="LAZY") 
*/ 
protected $nation; 
} 

ApiKey.php

/** 
* ApiKey 
* 
* @ORM\Table("api_key") 
* @ORM\Entity 
* @ORM\Entity(repositoryClass="AppBundle\Repository\ApiKeyRepository") 
*/ 
class ApiKey 
{ 
/** 
* @var string 
* @ORM\Column(name="api_key", unique=true, type="string", length=225) 
*/ 
protected $api_key; 

/** 
* @var \DateTime 
* @ORM\Column(name="expires", type="datetime") 
*/ 
protected $expires; 

/** 
* @ORM\OneToOne(targetEntity="User", inversedBy="apiKey") 
* @ORM\JoinColumn(name="user_id", referencedColumnName="id") 
* @ORM\Id 
*/ 
protected $user; 
} 

Answer 1

类的appbundle \实体\ ApiKey没有字段或协会命名user_ID的，但它有一个user字段。

所以，您的查询应该是这样的：

$result = $this->createQueryBuilder('a') 
       ->select(['a','u']) 
       ->leftJoin('a.user', u) 
       ->where('a.api_key = :api_key') 
       ->setParameter('api_key', $key) 
       ->getQuery() 
       ->getArrayResult();