获取实体中的当前登录用户 [英] Get current logged in user in entity

问题描述

我想在 n：m 关系中为实体创建一些虚拟属性。

I want to create some virtual properties for an entity in an n:m relation.

我有 User ， Achievment 和 AchievementUser 实体。用户在成就中拥有的值存储在实体 AchievementUser 中的字段 value 中。

I have an User, an Achievment and an AchievementUser entity. The value an user have in an Achievement is stored in the field value in the entity AchievementUser.

User -------- 1:n -------- AchievementUser -------- n:1 -------- Achievement
name:String                value:Integer                         name:String
[...]                                                            [...]

现在，我想使用成就本身返回用户在成就中拥有的值。所以我需要一个虚拟属性和 Achievement 实体中的方法 getValue（），但是要获取相应的 AchievementUser 对象，我需要当前登录用户的ID。

Now I want to return the value an user have in an achievement with the achievement itself. So I need a virtual property and a method getValue() in the Achievement entity, but to get the corresponding AchievementUser object, I need the ID of the current logged in user.

如何获取此信息？还是有其他可能获得成就的用户价值？谢谢您的帮助！

How can I get this? Or is there an other possiblility to get the user value for an achievement? Thanks for your help!

编辑：我只有一个基于API的应用程序。仅序列化程序执行Getter方法。这是我的序列化器文件的内容：

I only have an API based application. Only the serializer executes the Getter method. Here is the content of my serializer file:

virtual_properties:
    getValue:
        serialized_name: value
        type: integer
        groups: ['achievement_default']

推荐答案

您可以在Achievement实体中实现一个方法，并将当前经过身份验证的用户从twig模板的控制器传递到该方法中。

You can implement a method in the Achievement entity and pass the current authenticated user into it from your controller of twig template.

use Doctrine\Common\Collections\Criteria;

// ...

/**
 * @return Collection
 */
public function getAchievementUsers(User $user)
{
    $criteria = Criteria::create()->where(Criteria::expr()->eq('user', $user));

    return $this->achievementUsers->matching($criteria);
}

如果使用JMS序列化程序，则可以添加虚拟字段并填充通过定义序列化侦听器并注入TokenStorage来检索当前经过身份验证的用户，使用getAchievementUsers方法将其与数据一起使用。

In case of using a JMS serializer, you can add a virtual field and fill it with data using getAchievementUsers method by defining a serialization listener and injecting the TokenStorage to retrieve the current authenticated user.

<?php

namespace AppBundle\Listener\Serializer;

...
use JMS\Serializer\GenericSerializationVisitor;
use JMS\Serializer\EventDispatcher\ObjectEvent;
use Symfony\Component\Security\Core\Authentication\Token\Storage\TokenStorage;

class AchievementSerializerListener
{
    /**
     * @var User
     */
    protected $currentUser;

    /**
     * @param TokenStorage $tokenStorage
     */
    public function __construct(TokenStorage $tokenStorage)
    {
        $this->currentUser = $tokenStorage->getToken()->getUser();
    }

    /**
     * @param ObjectEvent $event
     */
    public function onAchievementSerialize(ObjectEvent $event)
    {
        if (!$this->currentUser) {
            return;
        }

        /** @var Achievement $achievement */
        $achievement = $event->getObject();

        /** @var GenericSerializationVisitor $visitor */
        $visitor = $event->getVisitor();

        $visitor->setData(
            'achievement_users',
            $achievement->getAchievementUsers($this->currentUser)
        );
    }
}

services.yml

services.yml

  app.listener.serializer.achievement:
        class: AppBundle\Listener\Serializer\AchievementSerializerListener
        arguments:
          - '@security.token_storage'
        tags: [ { name: jms_serializer.event_listener, event: serializer.post_serialize, class: AppBundle\Entity\Achievement, method: onAchievementSerialize } ]

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