php - 在 Symfony 中注册特定的表单类型 - Compiler Pass

Question

我已经创建了一个特定的表单类型:OrganizationsType。此类型在构造函数中采用 3 个参数。 tokenStorage (TokenStorageInterface)、Router(Router) 和我的类中的一个接口(interface)对象 (PreparatorInterface)。这个表单在一个 bundle 中，我希望每个人都通过实现 PreparatorInterface 来创建自己的 Preparator。

所以我想创建一个 Compiler pass，它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一个这样的表单时

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

我该怎么做？

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

---

期权值(value)

---

服务 ID OrganizationSelectorBundle\Form\OrganizationsForm
类 OrganizationSelectorBundle\Form\OrganizationsForm
标签 form.type
公开号
合成号
懒惰不
共享是
摘要号
Autowiring 是
自动配置是

---

当我添加 dump('Did I pass here') 时，我传递了我的编译器传递；死();

当我启动 bin/console debug:container --tag form.type 以列出所有带有标签 form.type 的服务时

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           
 --------------------------------------------------------------- ------- --------------------------------------------------------------- 

Answer 1

我找到了比使用 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