问题描述
我正在尝试使用 Sonata Admin Bundle 3 版本显示预览图像，但我无法做到。我在 RecipeAdmin.PHP 中收到此错误：无法加载类型“文件”：类不存在。
RecipeAdmin.PHP
<?PHP
declare(strict_types=1);
namespace App\Admin;
use Sonata\AdminBundle\Admin\AbstractAdmin;
use Sonata\AdminBundle\Datagrid\DatagridMapper;
use Sonata\AdminBundle\Datagrid\ListMapper;
use Sonata\AdminBundle\Form\FormMapper;
use Sonata\AdminBundle\Show\ShowMapper;
use Sonata\Form\Type\CollectionType;
use Sonata\AdminBundle\Form\Type\ModelListType;
final class RecipeAdmin extends AbstractAdmin
{
protected function configureDatagridFilters(DatagridMapper $datagridMapper): void
{
$datagridMapper
->add('title',null,['label' =>'Título'])
->add('image',['label' =>'Imagen'])
->add('description',['label' =>'Descripción'])
->add('score',['label' =>'Puntuación'])
->add('visible')
;
}
protected function configureListFields(ListMapper $listMapper): void
{
$listMapper
->add('id')
->add('user',CollectionType::class,['label' =>'Usuario'])
->add('title',['label' =>'Título'])
->add('image',['label' =>'Imagen'])
->add('description',['label' =>'Descripción'])
->add('score',['label' =>'Puntuación'])
->add('visible',['label' =>'Visible'])
->add('_action',[
'label' => 'Acciones','actions' => [
'show' => [],'edit' => [],'delete' => [],],]);
}
protected function configureFormFields(FormMapper $formMapper): void
{
$image = $this->getSubject();
// use $fileFormOptions so we can add other options to the field
$fileFormOptions = ['@R_404_3889@' => false];
if ($image && ($webPath = $image->getimage())) {
// get the request so the full path to the image can be set
$request = $this->getRequest();
$fullPath = $request->getBasePath().'/'.$webPath;
// add a 'help' option containing the preview's img tag
$fileFormOptions['help'] = '<img src="'.$fullPath.'" class="admin-preview"/>';
$fileFormOptions['help_html'] = true;
}
$formMapper
->add('title','file',$fileFormOptions)
->add('description',['label' =>'Puntuación'])
->add('visible')
;
}
protected function configureShowFields(ShowMapper $showMapper): void
{
$showMapper
->add('title',['label' =>'Descripción'])
->add('user',CollectionType::class)
->add('score',['label' =>'Puntuaciones'])
->add('visible')
;
}
}
Recipe.PHP 实体
<?PHP
namespace App\Entity;
use App\Repository\RecipeRepository;
use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\Common\Collections\Collection;
use Symfony\Component\HttpFoundation\File\UploadedFile;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity(repositoryClass=RecipeRepository::class)
*/
class Recipe
{
public function __construct() {
$this->categories = new \Doctrine\Common\Collections\ArrayCollection();
}
/**
* @ORM\Id
* @ORM\GeneratedValue
* @ORM\Column(type="integer")
*/
private $id;
/**
* @ORM\Column(type="string",length=255)
*/
private $title;
/**
* @ORM\Column(type="string",length=255)
*/
private $image;
/**
* @ORM\Column(type="string",length=255)
*/
private $description;
/**
* @ORM\ManyToOne(targetEntity="User",inversedBy="recipes")
* @ORM\JoinColumn(nullable=false)
*/
private $user;
/**
* @ORM\Column(type="integer",nullable=true)
*/
private $score;
/**
* @ORM\Column(type="boolean")
*/
private $visible;
/**
* @ORM\ManyToMany(targetEntity="Category",inversedBy="recipe",cascade={"persist"})
* @ORM\JoinTable(name="recipes_categories")
*/
private $categories;
public function getId(): ?int
{
return $this->id;
}
public function setId(int $id)
{
$this->id = $id;
return $this;
}
public function getTitle(): ?string
{
return $this->title;
}
public function setTitle(string $title): self
{
$this->title = $title;
return $this;
}
public function getimage(): ?string
{
return $this->image;
}
public function setimage(string $image): self
{
$this->image = $image;
return $this;
}
public function getDescription(): ?string
{
return $this->description;
}
public function setDescription(string $description): self
{
$this->description = $description;
return $this;
}
public function getUser()
{
return $this->user;
}
public function setUser($user)
{
$this->user = $user;
return $this;
}
public function getscore(): ?int
{
return $this->score;
}
public function setscore(?int $score): self
{
$this->score = $score;
return $this;
}
public function getVisible(): ?bool
{
return $this->visible;
}
public function setVisible(bool $visible): self
{
$this->visible = $visible;
return $this;
}
public function getCategories()
{
return $this->categories;
}
public function setCategories($categories)
{
$this->categories = $categories;
return $this;
}
public function __toString()
{
return $this->getTitle();
}
public function addCategory(Category $category): self
{
if (!$this->categories->contains($category)) {
$this->categories[] = $category;
}
return $this;
}
public function removeCategory(Category $category): self
{
$this->categories->removeElement($category);
return $this;
}
}
这是有关如何操作的链接：https://symfony.com/doc/current/bundles/SonataAdminBundle/cookbook/recipe_image_previews.html
在文档中解释说我必须使用“文件”类型的字段，但是当我在我的项目中使用它时不会运行。
解决方法
这是文档中的错误，您应该使用 file 而不是 FileType::class 并添加：
use Symfony\Component\Form\Extension\Core\Type\FileType;
$formMapper
->add('title',null,['label' =>'Título'])
->add('image',FileType::class,$fileFormOptions)
您仍然会遇到错误，例如：
The form's view data is expected to be an instance of class Symfony\Component\HttpFoundation\File\File,but is a(n) string. You can avoid this error by setting the "data_class" option to null or by adding a view transformer that transforms a(n) string to an instance of Symfony\Component\HttpFoundation\File\File.
在说明书中，他们告诉您创建一个图像实体，因此您应该添加它并按照所有步骤操作： https://sonata-project.org/bundles/admin/master/doc/cookbook/recipe_file_uploads.html
我建议您不要遵循这本说明书，而应该安装和使用 sonata media，它的集成更容易，而且它具有一些不错的功能，例如为您的上传制作不同的格式。