Php 在Syfmony-编译器过程中注册特定的表单类型

我创建了一个特定的表单类型：

OrganizationsType

。此类型在构造函数中接受3个参数。令牌存储（

TokenStorageInterface

）、路由器（

Router

）和我的类（

PreparatorInterface

）中的接口对象。这个表单在一个包中，我希望每个人通过实现

PreparatorInterface

来创建自己的

Preparator

所以我想创建一个编译器过程，可以用所有参数注册这个FormType。我试试这个：

$organizationListPreparator = $container->findTaggedServiceIds(self::TAG);

    if (empty($organizationListPreparator)) {
        throw new \Exception('CoffreoProOrganizationSelectorBundle need a preparator. Check README.');
    }

$container->register(OrganizationsForm::class, OrganizationsForm::class)
        ->addArgument(new Reference('security.token_storage'))
        ->addArgument(new Reference(key($organizationListPreparator)))
        ->addArgument(new Reference('router'))
        ->setAutoconfigured(true)
        ->setAutowired(true)
        ->setAbstract(true)
        ->addTag('form.type');

但是当我尝试实例化这样的表单时

$organizationsForm = $this->createForm(OrganizationsForm::class);

我怎么能做那样的事

编辑 我收到此错误消息：

Too few arguments to function OrganizationSelectorBundle\Form\OrganizationsForm::__construct(), 0 passed in /var/www/myProject/vendor/symfony/form/FormRegistry.php on line 92 and exactly 3 expected

[2018-12-18 16:34:38] request.CRITICAL: Uncaught PHP Exception Symfony\Component\Debug\Exception\FatalThrowableError: "Too few arguments to function OrganizationSelectorBundle\Form\OrganizationsForm::__construct(), 0 passed in /var/www/myProject/vendor/symfony/form/FormRegistry.php on line 92 and exactly 3 expected" at /var/www/myProject/vendor/organization-selector-bundle/Form/OrganizationsForm.php line 58 {"exception":"[object] (Symfony\\Component\\Debug\\Exception\\FatalThrowableError(code: 0): Too few arguments to function OrganizationSelectorBundle\\Form\\OrganizationsForm::__construct(), 0 passed in /var/www/myProject/vendor/symfony/form/FormRegistry.php on line 92 and exactly 3 expected at /var/www/myProject/vendor/organization-selector-bundle/Form/OrganizationsForm.php:58)"} []

编辑 当我启动

bin/console deb:container OrganizationForm

服务“OrganizationSelectorBundle\Form\OrganizationsForm”的信息

---

期权价值

---

服务ID OrganizationSelectorBundle\Form\OrganizationsForm
类组织SelectorBundle\Form\OrganizationsForm
标签表单类型
公共编号
合成no
懒惰不
共享是
摘要否
自动连线是
自动配置“是”

---

当我添加dump时，我传递了我的编译器传递（“我在这里传递了吗”）；模具（）

当我启动

bin/console debug:container--tag form.type

以列出带有tag

form.type的所有服务时

Symfony Container Services Tagged with "form.type" Tag
======================================================

 --------------------------------------------------------------- ------- --------------------------------------------------------------- 
  Service ID                                                      alias   Class name                                                     
 --------------------------------------------------------------- ------- --------------------------------------------------------------- 
  App\Form\StaffingCustomerType     App\Form\StaffingCustomerType                                  
  App\Form\UserType                 App\Form\UserType                                              
 OrganizationSelectorBundle\Form\OrganizationsForm      OrganizationSelectorBundle\Form\OrganizationsForm           
 --------------------------------------------------------------- ------- --------------------------------------------------------------- 

我找到了一个比使用CompilerPass更好的解决方案

我直接在yaml中使用服务定义，并将特定服务作为第二个参数传递。然后，我使用别名为此服务id提供默认服务：

OrganizationSelectorBundle\Form\OrganizationsForm:
    arguments:
      - '@security.token_storage'
      - '@coffreo_organization_bundle.organization_list_preparator'
      - '@router'
    tags: [form.type]
    public: false

coffreo_organization_bundle.organization_list_preparator:
    class: Coffreo\Pro\OrganizationSelectorBundle\Preparator\AllOrganizationListPreparator
    arguments: ['@translator']
    public: false

考虑用错误消息的相关部分更新你的问题（如果有的话）。是否已将OrganizationsForm从autowire中排除？你为什么把它抽象化？看起来你根本不需要这个，只要你要求“每个人”都有别名PreparatorInterface.Good。现在，使用“bin/console debug:container”验证您没有OrganizationsForm服务，并验证是否正在使用简单的die语句调用编译器传递代码。