Symfony/FOS-将用户ID变量传递给表单类型 [英] Symfony/FOS - pass the user id variable to a form Type

问题描述

我尝试从CategoryType表单中设置字段作者"的值.我希望它是使用FOS软件包登录的当前用户的用户ID.

I try to set the value of a field "author" from a CategoryType form. I want it to be the user id from the current user logged in with FOS bundle.

我的CategoryType表单:

namespace My\CategoryBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;

class CategoryType extends AbstractType
{
    private $userId;

    public function __construct(array $userId)
    {
        $this->userId = $userId;
    }

     /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('title')
            ->add('author')
            ->add('content')
        ;
    }

    /**
     * @param OptionsResolverInterface $resolver
     */
    public function setDefaultOptions(OptionsResolverInterface $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'My\CategoryBundle\Entity\Category',
            'auteur' => $this->userId
        ));
    }

    /**
     * @return string
     */
    public function getName()
    {
        return 'my_categorybundle_category';
    }
}

和我的控制器操作:

public function addAction()
{
    $category = new Category;
    $user = $this->get('security.context')->getToken()->getUser(); 
    $userId = $user->getId();

    $form = $this->get('form.factory')->create(new CategoryType(), array( 'author' => $userId));

    $request = $this->get('request');
    if ($request->getMethod() == 'POST') {
        $form->bind($request);

        if ($form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->persist($category);
            $em->flush();

        return $this->redirect($this->generateUrl('mycategory_voir',
            array('id' => $category->getId())));
        }
    }
    return $this->render('MyCategoryBundle:Category:add.html.twig',
        array(
            'form' => $form->createView(),
        ));
}

我在执行操作时捕获了此错误:

I catch this error while running the action :

可捕获的致命错误:传递给My \ CategoryBundle \ Form \ CategoryType :: __ construct()的参数1必须为数组，没有给出，在第55行的/My/CategoryBundle/Controller/CategoryController.php中调用并在/中定义My/CategoryBundle/Form/CategoryType.php第13行

Catchable Fatal Error: Argument 1 passed to My\CategoryBundle\Form\CategoryType::__construct() must be an array, none given, called in /My/CategoryBundle/Controller/CategoryController.php on line 55 and defined in /My/CategoryBundle/Form/CategoryType.php line 13

已经不是我要传递给表单的数组了吗?

Isn't it already an array that I am passing to the form ?

推荐答案

您的问题就在这行

$form = $this->get('form.factory')->create(new CategoryType(), array( 'author' => $userId));

您不满足My\CategoryBundle\FormCategoryType::__construct()的合同.在这里，让我们换一种方式看.

You're not satisfying the contract for My\CategoryBundle\FormCategoryType::__construct(). Here, let's look at it another way.

$form = $this->get('form.factory')->create(
    new CategoryType(/* You told PHP to expect an array here */)
  , array('author' => $userId)
);

作为Symfony\Component\Form\FormFactory::create()的第二个参数发送的数组最终将作为$options数组My\CategoryBundle\Form\CategoryType::buildForm()

The array that you send as the 2nd argument to Symfony\Component\Form\FormFactory::create() is what is ultimately injected as $options array My\CategoryBundle\Form\CategoryType::buildForm()

如我所见，您可以通过几种不同的方式来解决此问题

As I see it, you have a few different ways to resolve this

1. 更新参数签名并调用My\CategoryBundle\FormCategoryType::__construct()以传递/接收整个用户对象(不仅是其ID)-请记住，此时您正在使用Doctrine关系，而不是较低级的外键他们映射到的地方)

1. Update the argument signature AND call for My\CategoryBundle\FormCategoryType::__construct() to pass/receive the entire user object (not just their id - remember that you're working with Doctrine relationships at this point, not the lower-level foreign keys that they map to)

namespace My\CategoryBundle\Form;

use My\CategoryBundle\Entity\User; /* Or whatver your User class is */
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;

class CategoryType extends AbstractType
{
    private $author;

    public function __construct( User $author )
    {
        $this->author = $author;
    }

和

$form = $this->get('form.factory')->create(
    new CategoryType(
      $this->get('security.context')->getToken()->getUser()
    )
);

不要将User注入类型的构造函数中，只需让选项来处理

Don't inject the User into the type's constructor, just let the options handle it

namespace My\CategoryBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;

class CategoryType extends AbstractType
{
    private $userId;

     /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('title')
            ->add('author', 'hidden', array('data'=>$options['author']))
            ->add('content')
        ;
    }

    /**
     * @param OptionsResolverInterface $resolver
     */
    public function setDefaultOptions(OptionsResolverInterface $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'My\CategoryBundle\Entity\Category'
        ));
    }

    /**
     * @return string
     */
    public function getName()
    {
        return 'my_categorybundle_category';
    }
}

甚至不用费心把作者放在表单中，让控制器来处理

Not even bother putting the author in the form and let the controller handle it

$form = $this->get('form.factory')->create(
    new CategoryType()
  , array('author' => $this->get('security.context')->getToken()->getUser() )
);

和

if ($request->getMethod() == 'POST') {
    $form->bind($request);

    if ($form->isValid()) {
        $category->setAuthor(
          $this->get('security.context')->getToken()->getUser()
        );
        $em = $this->getDoctrine()->getManager();
        $em->persist($category);
        $em->flush();

    return $this->redirect($this->generateUrl('mycategory_voir',
        array('id' => $category->getId())));
    }
}

将您的表单类型转换为服务，并使用DI容器注入安全性上下文

Turn your form type into a service and user the DI Container to inject the security context

app/config/config.yml

services:
  form.type.my_categorybundle_category:
    class: My\CategoryBundle\Form\CategoryType
    tags:
      - {name: form.type, alias: my_categorybundle_category}
    arguments: ["%security.context%"]

更新您的类型以接收安全上下文

Update your type to receive the security context

namespace My\CategoryBundle\Form;

use Symfony\Component\Security\Core\SecurityContext;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;

class CategoryType extends AbstractType
{
    private $author;

    public function __construct( SecurityContext $security )
    {
        $this->author = $security->getToken()->getUser();
    }

然后在您的控制器中，使用服务名称创建表单

Then in your controller, create the form with its service name

$form = $this->get('form.factory')->create('my_categorybundle_category');

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