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Doctrine 2 - 实体通过密钥持有者表收集了另一个实体

[英]Doctrine 2 - entity have collection of another entities via key holder table

I want to implement this type of relationship: 我想实现这种类型的关系:

在此输入图像描述

Class SimpleUser must have a "languages" class property which should contain list of all SimpleLanguage-s that are binded to the current user in the UserLanguages table. 类SimpleUser必须具有“languages”类属性,该属性应包含绑定到UserLanguages表中当前用户的所有SimpleLanguage-s的列表。

Please look at the classes: 请查看课程:

public class SimpleUser {

    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @var string
     *
     * @ORM\Column(name="name", type="string", length=50)
     */
    private $name;

    // what do I type here? I want to have SimpleLanguage[] here
    private $languages;
}

 public class SimpleLanguage {

    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @var string
     *
     * @ORM\Column(name="name", type="string", length=50)
     */
    private $name;
}

 public class UserLanguages {

    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;


/**
 * @var integer
 * @ORM\Column(name="user_id", type="integer")
 */
private $userId;

/**
 * @var integer
 * @ORM\Column(name="language_id", type="integer")
 */
private $languageId;

/**
 * @var SimpleUser
 *
 * @ManyToOne(targetEntity="SimpleUser")
 * @JoinColumn(name="userId", referencedColumnName="id")
 */
private $user;

/**
 * @var SimpleLanguage
 *
 * @ManyToOne(targetEntity="SimpleLanguage")
 * @JoinColumn(name="languageId", referencedColumnName="id")
 */
private $language;
}

First delete $userId; 首先删除$ userId; and $languageId from your class UserLanguages ! 你的班级UserLanguages的$ languageId They are useless, these are properties $user and $language which manage the id of the entity in the database ! 它们是无用的,这些是属性$ user和$ language,它们管理数据库中实体的id!

Here how to manage languages property in SImpleUser : 这里是如何管理SImpleUser中的语言属性:

use Doctrine\Common\Collections\ArrayCollection;

// ...

class SimpleUser
{
    //...

    /**
     * @ORM\OneToMany(targetEntity="UserLanguages", mappedBy="user")
     */
    protected $languages;

    public function __construct()
    {
        $this->languages = new ArrayCollection();
    }
}

Similar in SimpleLanguage : 类似于SimpleLanguage:

use Doctrine\Common\Collections\ArrayCollection;

// ...

class SimpleLanguage
{
    //...

    /**
     * @ORM\OneToMany(targetEntity="UserLanguages", mappedBy="language")
     */
    protected $languages;

    public function __construct()
    {
        $this->languages = new ArrayCollection();
    }
}

EDIT: 编辑:

I want to have not collection of "UserLanguages" but directly collection of "SimpleLanguage" from the "user_languages" table 我想没有“UserLanguages”的集合,而是直接从“user_languages”表中收集“SimpleLanguage”

Ok currently this is not possible, I think want a ManyToMany relationship instead of a third entity UserLanguages (your question is not very clear) ! 好的,目前这是不可能的,我认为想要一个ManyToMany关系而不是第三个实体UserLanguages(你的问题不是很清楚)!

Delete the useless entity UserLanguages, and then in SimpleUser : 删除无用的实体UserLanguages,然后在SimpleUser中:

use Doctrine\Common\Collections\ArrayCollection;

// ...

class SimpleUser
{
    //...

    /**
     * @ORM\ManyToMany(targetEntity="SimpleLanguage", mappedBy="users")
     */
    protected $languages;

    public function __construct()
    {
        $this->languages = new ArrayCollection();
    }
}

And SimpleLanguage : 而SimpleLanguage:

use Doctrine\Common\Collections\ArrayCollection;

// ...

class SimpleLanguage
{
    //...

    /**
     * @ORM\ManyToMany(targetEntity="SimpleUser", mappedBy="languages")
     * @JoinTable(name="UserLanguages")
     */
    protected $users;

    public function __construct()
    {
        $this->users = new ArrayCollection();
    }
}

It will result a third table UserLanguages having user_id and language_id ! 这将导致第三个表UserLanguages具有user_id和language_id!

You'll be able from user class get the list of all languages linked with one use : 您将能够从用户类获取与一次使用链接的所有语言的列表:

$languages = $user->getLanguages();

Inversely for a language : 反向语言:

$users = $language->getUsers();

You are looking for a uni-directional one-to-many relationship with a join-table which is described in doctrine with the @ManyToMany annotation. 您正在寻找与连接表的单向一对多关系,该连接表在学说中使用@ManyToMany注释进行描述。 This can lead to confusion. 这可能会导致混淆。

You don't even need the glueing UserLanguages class at all ... doctrine will take care of the table for you. 你根本不需要粘合UserLanguages类... doctrine会为你处理这个表。

SimpleUser SimpleUser

/**
 * @ORM\ManyToMany(targetEntity="SimpleLanguage", inversedBy="user")
 * @ORM\JoinTable(name="UserLanguages",
 *      joinColumns={@ORM\JoinColumn(name="user_id", referencedColumnName="id")},
 *      inverseJoinColumns={@ORM\JoinColumn(name="language_id", referencedColumnName="id", unique=true)}
 * )
 */
 protected $languages;

SimpleLanguage SimpleLanguage

/**
 * @ORM\ManyToMany(targetEntity="SimpleUser", mappedBy="languages")
 */
 protected $user;

now doctrine will create the UserLanguages table for you and keep the id's ( user_id , language_id ) in there. 现在doctrine将为您创建UserLanguages表并保留id(user_id,language_id)。

Read more on association mappings with a join-table here . 此处阅读有关联接表的关联映射的更多信息

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