小编Cyb*_*101的帖子

Symfony 4注册表第二次提交"已经有一个活跃的交易"

我已经创建了一个用户注册表,就像在文档中一样.稍后我想添加CSRF.从那以后它再也不能正常工作了.我安装并删除了包. https://symfony.com/doc/current/doctrine/registration_form.html

composer require security-csrf
composer remove security-csrf
rm -rf vendor
rm -rf var/cache/*
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这没有用.

当我填写表格并提交.如果用户已保存.我第二次,我收到下面的错误消息.

只有当我在浏览器中删除cookie"PHPSESSID"时,我才能再次发送表单.之后再次出现同样的问题.

PDOException:
There is already an active transaction

  at vendor/doctrine/dbal/lib/Doctrine/DBAL/Connection.php:1249
  at PDO->beginTransaction()
     (vendor/doctrine/dbal/lib/Doctrine/DBAL/Connection.php:1249)
  at Doctrine\DBAL\Connection->beginTransaction()
     (vendor/doctrine/orm/lib/Doctrine/ORM/UnitOfWork.php:376)
  at Doctrine\ORM\UnitOfWork->commit(null)
     (vendor/doctrine/orm/lib/Doctrine/ORM/EntityManager.php:358)
  at Doctrine\ORM\EntityManager->flush()
     (src/Controller/RegistrationController.php:40)
  at App\Controller\RegistrationController->registerAction(object(Request), object(UserPasswordEncoder))
     (vendor/symfony/http-kernel/HttpKernel.php:149)
  at Symfony\Component\HttpKernel\HttpKernel->handleRaw(object(Request), 1)
     (vendor/symfony/http-kernel/HttpKernel.php:66)
  at Symfony\Component\HttpKernel\HttpKernel->handle(object(Request), 1, true)
     (vendor/symfony/http-kernel/Kernel.php:190)
  at Symfony\Component\HttpKernel\Kernel->handle(object(Request))
     (public/index.php:37)
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我试着调试一下.交易在某处激活,然后不再激活.我找不到问题.有人能帮助我吗?

的symfony /配置/ services.yaml

services:
    Symfony\Component\HttpFoundation\Session\Storage\Handler\PdoSessionHandler:
        arguments:
            - !service { class: PDO, factory: 'database_connection:getWrappedConnection' }
            - { db_table: session, db_username: username, db_password: password }
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php registration symfony

5
推荐指数
1
解决办法
1192
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将php symfony中的多个模型从json解码为对象

我想将 json 转换为对象/模型。如果 json 只是一维的,它可以完美地工作。

但是如果它是多维的,则只转换外部(用户),而不转换内部(公司),这仍然是一个数组。你能帮我解决这个问题吗?

模型:

<?php
namespace AppBundle;

class Company {
    /**
     * @var string
     */
    protected $companyName = '';

    /**
     * @return string
     */
    public function getCompanyName()
    {
        return $this->companyName;
    }

    /**
     * @param string $companyName
     * @return void
     */
    public function setCompanyName($companyName)
    {
        $this->companyName = $companyName;
    }
}

class User {
    /**
     * @var \AppBundle\Company
     */
    protected $company = null;

    /**
     * @var string
     */
    protected $username = '';

    /**
     * @return \AppBundle\Company
     */
    public function …
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php json model decode symfony-3.1

2
推荐指数
1
解决办法
3136
查看次数

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php ×2

decode ×1

json ×1

model ×1

registration ×1

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symfony-3.1 ×1