blocks|key|762446|text|首先，您需要在控制器操作中从会话中请求用户的用户名，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|762447|$username=$this->get('security.context')->getToken()->getUser()->getUserName();|code-block|syntax|javascript|762448|然后对db执行查询，并使用常规dql获取对象，如下所示|762449|$em+=+$this->get('doctrine.orm.entity_manager');++++
"SELECT+u+FROM+Acme\AuctionBundle\Entity\User+u+where+u.username=".$username;
$q=$em->createQuery($query);
$user=$q->getResult();|762450|$user现在应该包含具有此用户名的用户(当然，您也可以使用其他字段)|762451|...but您必须首先配置/app/config/security.yml配置，以便为您的安全提供商使用适当的字段，如下所示：|762452|security:
+provider:
++example:
+++entity:+{class+Acme\AuctionBundle\Entity\User,+property:+username}|762453|希望这能有所帮助！|762454|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Well, first you need to request the username of the user from the session in your controller action like this:

<pre><code>$username=$this-&gt;get('security.context')-&gt;getToken()-&gt;getUser()-&gt;getUserName();
</code></pre>

then do a query to the db and get your object with regular dql like

<pre><code>$em = $this-&gt;get('doctrine.orm.entity_manager'); 
"SELECT u FROM Acme\AuctionBundle\Entity\User u where u.username=".$username;
$q=$em-&gt;createQuery($query);
$user=$q-&gt;getResult();
</code></pre>

the $user should now hold the user with this username ( you could also use other fields of course)

...but you will have to first configure your /app/config/security.yml configuration to use the appropriate field for your security provider like so:

<pre><code>security:
 provider:
 example:
 entity: {class Acme\AuctionBundle\Entity\User, property: username}
</code></pre>

hope this helps!

blocks|key|545896|text|这个帖子有点旧，但我认为这可能会节省一些人的时间……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|545897|我遇到了与原始问题相同的问题，类型显示为Symfony\Component\Security\Core\User\User|545898|最终证明，我是使用内存中的用户登录的|545899|我的security.yml看起来像这样|545900|security:
++++providers:
++++++++chain_provider:
++++++++++++chain:
++++++++++++++++providers:+[in_memory,+fos_userbundle]
++++++++fos_userbundle:
++++++++++++id:+fos_user.user_manager
++++++++in_memory:
++++++++++++memory:
++++++++++++++++users:
++++++++++++++++++++user:++{+password:+userpass,+roles:+[+'ROLE_USER'+]+}
++++++++++++++++++++admin:+{+password:+adminpass,+roles:+[+'ROLE_ADMIN',+'ROLE_SONATA_ADMIN'+]+}|code-block|syntax|javascript|545901|核心用户类型始终是Symfony\Component\Security\+in_memory+\+user+\+user如果要使用自己的实体，请使用该提供程序的用户登录。|545902|谢谢，hj|545903|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

The thread is a bit old but i think this could probably save someone's time ...

I ran into the same problem as the original question, that the type is showed as
Symfony\Component\Security\Core\User\User

It eventually turned out that i was logged in using an in memory user

my security.yml looks something like this

<pre><code>security:
 providers:
 chain_provider:
 chain:
 providers: [in_memory, fos_userbundle]
 fos_userbundle:
 id: fos_user.user_manager
 in_memory:
 memory:
 users:
 user: { password: userpass, roles: [ 'ROLE_USER' ] }
 admin: { password: adminpass, roles: [ 'ROLE_ADMIN', 'ROLE_SONATA_ADMIN' ] }
</code></pre>

the in_memory user type is always Symfony\Component\Security\Core\User\User if you want to use your own entity, log in using that provider's user.

Thanks,
hj

blocks|key|545926|text|$this->container->get('security.token_storage')->getToken()->getUser();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|545927|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>$this-&gt;container-&gt;get('security.token_storage')-&gt;getToken()-&gt;getUser();
</code></pre>

blocks|key|762635|text|在Symfony版本PHP5和>=+>=+8.0中，您可以使用属性来提示经过身份验证的用户，但只能在控制器中输入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|762636|public+function+indexAction(#[CurrentUser]+User+$user):+Response
{
+++++//+do+something
}|code-block|syntax|javascript|762637|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In Symfony version &gt;= 5 and PHP &gt;= 8.0 you can type hint the authenticated user using Attributes, only in controllers though:
<pre class="lang-php prettyprint-override"><code>public function indexAction(#[CurrentUser] User $user): Response
{
 // do something
}
</code></pre>

I am using the Symfony security setup. Everything works fine, but I don't know how to do one important thing:

In twig, I can reach the current user's info by doing:

<pre><code>Welcome, {{ app.user.username }}
</code></pre>

or similar

How do I access this same information in the Controller? Specifically, I want to get the current user entity so I can store it relationally in another entity (one-to-one mapping).

I was really hoping it'd be

<pre><code>$this-&gt;get('security.context')-&gt;getToken()-&gt;getUser()
</code></pre>

but that doesn't work. It gives me a class of type

<pre><code>Symfony\Component\Security\Core\User\User
</code></pre>

and I want one of type

<pre><code>Acme\AuctionBundle\Entity\User
</code></pre>

which is my entity....

How do I get the entity that represents the current user in Symfony2?

身份验证

我正在使用Symfony安全设置。一切都很好，但我不知道如何做一件重要的事情：在twig中，我可以通过执行以下操作来获取当前用户的信息：Welcome, {{ app.user.username }}或类似的如何在控制器中访问相同的信息？具体地说，我希望获得当前的用户实体，这样我就可以将其以关系的方式存储在另一个实体中...

问如何在Symfony2中获取表示当前用户的实体？
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