![](/img/trans.png)
[英]CKEditor not showing with Sonata Formatter (Sonata Admin Bundle)
[英]Showing image previews Sonata Admin Bundle Could not load type "file"
我正在嘗試使用 Sonata Admin Bundle 3 版本顯示預覽圖像,但我做不到。 我在 RecipeAdmin.php 中收到此錯誤:無法加載類型“文件”:class 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',null,['label' =>'Imagen'])
->add('description',null,['label' =>'Descripción'])
->add('score',null,['label' =>'Puntuación'])
->add('visible')
;
}
protected function configureListFields(ListMapper $listMapper): void
{
$listMapper
->add('id')
->add('user', CollectionType::class,['label' =>'Usuario'])
->add('title',null,['label' =>'Título'])
->add('image',null,['label' =>'Imagen'])
->add('description',null,['label' =>'Descripción'])
->add('score',null,['label' =>'Puntuación'])
->add('visible',null,['label' =>'Visible'])
->add('_action', null, [
'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
$fileFormOptions = ['required' => false];
if ($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',null,['label' =>'Título'])
->add('image', 'file', $fileFormOptions)
->add('description',null,['label' =>'Descripción'])
->add('score',null,['label' =>'Puntuación'])
->add('visible')
;
}
protected function configureShowFields(ShowMapper $showMapper): void
{
$showMapper
->add('title',null,['label' =>'Título'])
->add('image',null,['label' =>'Imagen'])
->add('description',null,['label' =>'Descripción'])
->add('user', CollectionType::class)
->add('score',null,['label' =>'Puntuaciones'])
->add('visible')
;
}
}
配方.php 實體
<?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;
}
}
這是有關如何操作的鏈接: https://symfony.com/doc/current/bundles/SonataAdminBundle/cookbook/recipe_image_previews.html
在文檔中解釋說我必須使用“文件”類型字段但是當我在我的項目中使用它時不會運行。
這是文檔中的錯誤,而不是file
,您應該使用FileType::class
並添加:
use Symfony\Component\Form\Extension\Core\Type\FileType;
$formMapper
->add('title',null,['label' =>'Título'])
->add('image', FileType::class, $fileFormOptions)
您仍然會遇到錯誤,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.
在食譜中,他們告訴您創建一個圖像實體,因此您應該添加它並按照所有步驟操作: https://sonata-project.org/bundles/admin/master/doc/cookbook/recipe_file_uploads.html
我建議您不要遵循這本食譜,而應該安裝和使用奏鳴曲媒體,集成更容易,並且它具有一些不錯的功能,例如為您的上傳制作不同的格式。
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