我正在使用 Gedmo Doctrine Extensions 将类别作为嵌套集来处理。我正在构建一个 REST API,并且我有一个创建类别的路线。我希望能够创建根类别或子类别。这是实体
<?php
namespace AppBundle\Entity;
use Gedmo\Mapping\Annotation as Gedmo;
use Doctrine\ORM\Mapping as ORM;
/**
* @Gedmo\Tree(type="nested")
* @ORM\Table(name="bo_categories")
* use repository for handy tree functions
* @ORM\Entity(repositoryClass="Gedmo\Tree\Entity\Repository\NestedTreeRepository")
*/
class Category
{
/**
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue
*/
private $id;
/**
* @ORM\Column(name="name", type="string", length=64)
*/
private $name;
/**
* @Gedmo\TreeLeft
* @ORM\Column(name="lft", type="integer")
*/
private $lft;
/**
* @Gedmo\TreeLevel
* @ORM\Column(name="lvl", type="integer")
*/
private $lvl;
/**
* @Gedmo\TreeRight
* @ORM\Column(name="rgt", type="integer")
*/
private $rgt;
/**
* @Gedmo\TreeRoot
* @ORM\Column(name="root", type="integer", nullable=true)
*/
private $root;
/**
* @Gedmo\TreeParent
* @ORM\ManyToOne(targetEntity="Category", inversedBy="children")
* @ORM\JoinColumn(name="parent_id", referencedColumnName="id", onDelete="CASCADE")
*/
private $parent;
/**
* @ORM\OneToMany(targetEntity="Category", mappedBy="parent")
* @ORM\OrderBy({"lft" = "ASC"})
*/
private $children;
public function getId()
{
return $this->id;
}
public function setName($name)
{
$this->name = $name;
}
public function getName()
{
return $this->name;
}
public function setParent(Category $parent = null)
{
$this->parent = $parent;
}
public function getParent()
{
return $this->parent;
}
}
这是表格
<?php
namespace AppBundle\Form\Type\Classification;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;
class CategoryFormType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('name')
->add('parent', 'entity', array(
'class' => 'AppBundle:Category',
'property' => 'id',
))
;
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults(array(
'data_class' => 'AppBundle\Entity\Category',
'csrf_protection' => false,
));
}
public function getName()
{
return 'api_category';
}
}
这是控制器
/**
* @Route("/create")
* @Security("has_role('ROLE_SUPER_ADMIN')")
* @Rest\View
*/
public function postCreateAction(Request $request)
{
$categoryManager = $this->get('app.manager.category');
$category = $categoryManager->createNew();
$form = $this->createForm(new CategoryFormType(), $category);
// $category->setParent(10);
$form->handleRequest($request);
if ($form->isValid()) {
$categoryManager->save($category);
return new Response('', 201);
} else {
return $this->view($form, 400);
}
}
如果我想创建一个子类别,它可以正常工作。但是,如果我想创建一个根类别而不删除表单中的“父”字段,我会收到此错误
An exception occurred while executing 'SELECT b0_.id AS id0, b0_.name AS name1, b0_.lft AS lft2, b0_.lvl AS lvl3, b0_.rgt AS rgt4, b0_.root AS root5, b0_.parent_id AS parent_id6 FROM bo_categories b0_ WHERE b0_.id IN (?)' with params [""]:\n\nSQLSTATE[22P02]: Invalid text representation: 7 ERROR: invalid input syntax for integer: "",
"class": "Doctrine\DBAL\DBALException
为什么我会收到此错误?表格中的“父”值不能为空/null吗?