Symfony 将对象持久化到数据库错误
我有一个错误:Symfony 将对象持久化到数据库错误,symfony,orm,doctrine,persistent,symfony4,Symfony,Orm,Doctrine,Persistent,Symfony4,我有一个错误: Expected argument of type "integer or null", "App\Entity\User" given 我不明白,我明白这个错误,但我不知道为什么会出现 以下是我的AddController.php文件: <?php namespace App\Controller\Tag; use App\Entity\Tag; use App\Form\Tag\AddType; use Symfony\Bundle\
Expected argument of type "integer or null", "App\Entity\User" given
我不明白,我明白这个错误,但我不知道为什么会出现
以下是我的AddController.php文件:
<?php
namespace App\Controller\Tag;
use App\Entity\Tag;
use App\Form\Tag\AddType;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\Config\Definition\Exception\Exception;
use Symfony\Component\HttpFoundation\Request;
class AddController extends Controller
{
public function add(Request $request)
{
$tag = new Tag();
$form = $this->createForm(AddType::class, $tag);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$entityManager = $this->getDoctrine()->getManager();
$pageInfo = $form->getData();
$description = $pageInfo['description'];
$name = $pageInfo['name'];
$guru = $pageInfo['guru_id'];
$createdTs = new \DateTime();
$tag->setApproved(false);
$tag->setDescription($description);
$tag->setName($name);
$tag->setGuruId((is_int($guru) ? $guru : null));
$tag->setCreatedTs($createdTs);
try {
$entityManager->persist($tag);
$entityManager->flush();
$this->addFlash('success', 'Tag Submitted for review! '. $guru);
} catch (Exception $e) {
$this->addFlash('danger', 'Something went skew-if. Please try again.');
}
return $this->redirectToRoute('tag_add');
}
return $this->render('tag/add.html.twig', array('form' => $form->createView()));
}
}
我的标记实体:
<?php
namespace App\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity(repositoryClass="App\Repository\TagRepository")
*/
class Tag
{
/**
* @ORM\Id()
* @ORM\GeneratedValue()
* @ORM\Column(type="integer")
*/
private $id;
/**
* @ORM\Column(type="string", length=125)
*/
private $name;
/**
* @ORM\Column(type="string", length=255, nullable=true)
*/
private $description;
/**
* @ORM\Column(type="boolean")
*/
private $approved;
/**
* @ORM\Column(type="datetime")
*/
private $created_ts;
/**
* @ORM\Column(type="datetime", nullable=true)
*/
private $last_edit_ts;
/**
* @ORM\Column(type="integer", nullable=true)
*/
private $guru_id;
public function getId()
{
return $this->id;
}
public function getName(): ?string
{
return $this->name;
}
public function setName(string $name): self
{
$this->name = $name;
return $this;
}
public function getDescription(): ?string
{
return $this->description;
}
public function setDescription(?string $description): self
{
$this->description = $description;
return $this;
}
public function getApproved(): ?bool
{
return $this->approved;
}
public function setApproved(bool $approved): self
{
$this->approved = $approved;
return $this;
}
public function getCreatedTs(): ?\DateTimeInterface
{
return $this->created_ts;
}
public function setCreatedTs(\DateTimeInterface $created_ts): self
{
$this->created_ts = $created_ts;
return $this;
}
public function getLastEditTs(): ?\DateTimeInterface
{
return $this->last_edit_ts;
}
public function setLastEditTs(\DateTimeInterface $last_edit_ts): self
{
$this->last_edit_ts = $last_edit_ts;
return $this;
}
public function getGuruId(): ?int
{
return $this->guru_id;
}
public function setGuruId(?int $guru_id): self
{
$this->guru_id = $guru_id;
return $this;
}
}
它仍然给出相同的错误,所以我真的不确定它是如何生成的。。。如何将EntityType下拉列表中的整数持久化到数据库中
谢谢您可能在AddType表单中有一个EntityType字段来处理guru_id属性,它正在使用setGuruId?int$guru_id方法自动映射用户提交的值实例。这就是为什么:
Expected argument of type "integer or null", "App\Entity\User" given
您有一些选项可以解决这个问题,但最好的方法可能是将guru_id映射更改为与用户的多通关系:
此外,我将删除以下代码:
$tag->setDescription($description);
$tag->setName($name);
$tag->setGuruId((is_int($guru) ? $guru : null));
因为表单组件在提交数据后已经完成了此过程
作为一种快速解决方法,您可以通过删除typehint来解决此问题:
public function setGuruId($guru_id): self
{
$this->guru_id = $guru_id instanceof User ? $guru_id->getId() : $guru_id;
return $this;
}
请将您的问题链接到您的标签实体。我认为你使用条令映射是错误的way@Mcsky它是我的make:entity自动生成的-还需要代码还是?寻找密码yep@Mcsky查看更新的问题古鲁是您的表中与标签相关的吗?如果是的话,这是一个映射原则的问题。我写的答案不是用户密钥,它们之间没有关系。正如您在实体中看到的,guru是integerI类型的一个简单字段。。。但就目前而言,并没有暗示guru正在接收用户id。应该先询问表单内容和完整的错误消息。。。OP可能遗漏了其他内容。@Preciel这可能是它^^^ guru不是一个用户,但它受…这是它indeedy:ty:
// Tag entity
/**
* @ORM\ManyToOne(targetEntity="App\Entity\User")
*/
private $guru;
public function getGuru(): ?User
{
return $this->guru;
}
public function setGuru(?User $guru): self
{
$this->guru = $guru;
return $this;
}
$tag->setDescription($description);
$tag->setName($name);
$tag->setGuruId((is_int($guru) ? $guru : null));
public function setGuruId($guru_id): self
{
$this->guru_id = $guru_id instanceof User ? $guru_id->getId() : $guru_id;
return $this;
}