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问题[symfony]

Martin Hope
Михаил Сазанов
Asked: 2023-01-06 20:53:20 +0000 UTC
  • 6

我在orm学说中创建实体，ID配置为

#[ORM\Id]
#[ORM\Column(type: Types::INTEGER)]
#[ORM\GeneratedValue]
private int $id;

, 实体有唯一键

#[ORM\Column(type: Types::STRING, unique: true)]
private string $username;

#[ORM\Column(type: Types::STRING, unique: true)]
private string $email;

问题是，如果您尝试使用相同的电子邮件或用户名在事务中添加第二条记录，尽管 INSERT 不会被执行并且异常将被捕获 "An exception occurred while executing a query: Duplicate entry 'username' for key 'users.username'"并完成$conn->rollBack()

, AUTO_INCREMENT 会增加。在下一次添加时（如果在此之前成功的 INSERT 返回 ID = 1），它 (ID) 将不是 2，而是 3。

如何正确实现正确的缩放和/或 AI 整形？

UPD：+ 添加了 prePersist

public function checkExistingRecords(array $fields, LifecycleEventArgs $args) {
        $obj = $args->getObject();
        $repository = $args->getObjectManager()->getRepository($obj::class)->findOneBy($fields);
        if($repository instanceof $obj) {
            throw new EntityAlreadyExistsException($obj::class);
        }

    }
symfony
Martin Hope
Михаил Ребров
Asked: 2022-07-30 15:48:35 +0000 UTC
  • 0

我想将我的.yaml-file 中的字段值留空。
只是在一个翻译中该字段必须填写，而在另一个翻译中则不能。

如果未定义此字段，则其值 ( ...title.3) 的路径将显示在前面。

一定有什么办法。

想要这样的东西：

title:
    1: String
    2: String2
    3:

因此问题是：
是否有可能以yaml某种方式在字段中设置一个空值，如果可以，如何设置？

php symfony
Martin Hope
Garibaldi
Asked: 2022-07-14 22:46:14 +0000 UTC
  • 0

有一个Offer实体

* @ApiResource(
 *     normalizationContext={"groups"={"offer:read"}, "swagger_definition_name"="Read"},
 *     denormalizationContext={"groups"={"offer:write"}, "swagger_definition_name"="Write"},
 *     itemOperations={
 *         "get",
 *         "put"={"security"="is_granted('ROLE_USER')"},
 *         "delete": {
 *             "security"="is_granted('ROLE_USER')"
 *         }
 *     },
 *     collectionOperations={
 *          "get",
 *          "post"={
 *              "security"="is_granted('ROLE_USER')"
 *          }
 *     },
 *     attributes={
 *          "order"={"id": "DESC"}
 *     },
 * )
class Offer
{
...

    /**
     * @var Collection страны оффера
     *
     * @ManyToMany(targetEntity=Country::class)
     * @JoinTable(name="offers_countries",
     *      joinColumns={@JoinColumn(name="offer_id", referencedColumnName="id")},
     *      inverseJoinColumns={@JoinColumn(name="country_id", referencedColumnName="id")}
     * )
     * @Assert\NotBlank
     * @Assert\NotNull
     * @Assert\Count(min=1)
     * @Groups({"offer:read", "offer:write"})
     * @ApiSubresource(maxDepth=1000)
     */
    private Collection $countries;

    /**
     * @return Country[]|ArrayCollection<Country>
     */
    public function getCountries()
    {
        return $this->countries;
    }

    /**
     * @param Country $country
     * @return $this
     */
    public function addCountry(Country $country): self
    {
        if (!$this->countries->contains($country)) {
            $this->countries->add($country);
        }

        return $this;
    }
  ...
}

使用诸如https://example.com/api/offers/1 "countries": ["/api/countries/10"] 之类的 PUT 请求 ，将 ID 为 10 的国家/地区添加到现有国家/地区。告诉我出了什么问题。谢谢

symfony doctrine2
Martin Hope
F. Tomas
Asked: 2022-09-06 03:54:10 +0000 UTC
  • 1

大家好。我正在新的 Symfony 6.1 上做一个小项目。我连接了设备，因此有必要对密码进行哈希处理。翻遍码头，发现 PasswordHasherInterface

<?php
namespace App\DataFixtures;

use App\Model\User\Entity\User\Email;
use App\Model\User\Entity\User\Id;
use App\Model\User\Entity\User\Role;
use App\Model\User\Entity\User\User;
use Doctrine\Bundle\FixturesBundle\Fixture;
use Doctrine\Persistence\ObjectManager;
use Symfony\Component\PasswordHasher\PasswordHasherInterface;

class UserFixture extends Fixture
{
    private PasswordHasherInterface $hasher;

    public function __construct(PasswordHasherInterface $hasher)
    {
        $this->hasher = $hasher;
    }
    
    public function load(ObjectManager $manager): void
    {
        $hash = $this->hasher->hash("password");

        $user = User::signUpByEmail(
            Id::next(),
            new \DateTimeImmutable(),
            new Email("admin@app.test"),
            $hash,
            "token"
        );

        $user->confirmSignUp();

        $user->changeRole(Role::admin());

        $manager->persist($user);
        $manager->flush();
    }
}

但我收到一个错误：

In DefinitionErrorExceptionPass.php line 54:
!!                                                                                 
!!    Cannot autowire service "App\DataFixtures\UserFixture": argument "$hasher"   
!!    of method "__construct()" references interface "Symfony\Component\PasswordH  
!!    asher\PasswordHasherInterface" but no such service exists. Did you create a  
!!    class that implements this interface?                                       
!!

我的 services.yaml 文件：

parameters:

services:
    # default configuration for services in *this* file
    _defaults:
        autowire: true      # Automatically injects dependencies in your services.
        autoconfigure: true # Automatically registers your services as commands, event subscribers, etc.

    # makes classes in src/ available to be used as services
    # this creates a service per class whose id is the fully-qualified class name
    App\:
        resource: '../src/'
        exclude:
            - '../src/DependencyInjection/'
            - '../src/Model/User/Entity/'
            - '../src/Kernel.php'

如何在 Symfony 6.1 中散列密码？

php symfony