Question

在Symfony admin中我有一个表单，其中第二个字段类型取决于所选的选择字段值。第二个字段可以是Symfony Url 字段类型，也可以是提供 sonata_type_model_list 字段类型的奏鸣曲。

我已经向My Bundle Controller创建了一个ajax请求，以返回包含所需字段的表单。

> /src/MyBundle/Controller/MyController.php

namespace MyBundle\Controller

use Sonata\AdminBundle\Controller\CRUDController;
use Symfony\Component\HttpFoundation\Request;
use Doctrine\ORM\Mapping\ClassMetadataInfo;
use Sonata\AdminBundle\Form\FormMapper;

class MyController extends CRUDController
{
    public function getFieldAction()
  {
    //getting the value of choice field
    $type = $this->get('request')->get('type'); 

    //sonata.admin.reference is a service name of ReferenceBundle admin class
        $fieldDescription = $this->admin->getModelManager()
         ->getNewFieldDescriptionInstance($this->admin->getClass(), 'reference');
        $fieldDescription->setAssociationAdmin($this->container->get('sonata.admin.reference'));
        $fieldDescription->setAdmin($this->admin);
        $fieldDescription->setAssociationMapping(array(
            'fieldName' => 'reference',
            'type' => ClassMetadataInfo::ONE_TO_MANY,
        ));

    // Getting form mapper in controller:
    $contractor = $this->container->get('sonata.admin.builder.orm_form');
    $mapper = new FormMapper($contractor, $this->admin->getFormBuilder(), $this->admin);

    $form_mapper = $mapper->add('reference', 'sonata_type_model_list', array(
            'translation_domain' => 'ReferenceBundle',
            'sonata_field_description' => $fieldDescription,
            'class' => $this->container->get('sonata.admin.reference')->getClass(),
            'model_manager' => $this->container->get('sonata.admin.reference')->getModelManager(),
            'label' => 'Reference',
            'required' => false,
        ));


    //@ToDo build $form from $form_mapper


    return $this->render('MyBundle:Form:field.view.html.twig', array(
        'form' => $form->createView(),
    ));
  }
}

我在Sonata \ AdminBundle \ Form \ FormMapper类中找不到任何方法来构建表单（似乎可以使用 create（）方法，但它只适用于常见的Symfony字段类型，不是Sonata表单字段类型，通常在Block或Admin类中生成。）

是否可以在Controller 中使用 Sonata \ AdminBundle \ Form \ FormMapper 来构建表单？或者是否有另一种方法可以使用Sonata表单字段类型在Controller 中构建表单？

Answer 1

您不应使用控制器，而应使用Sonata Admin Service。

Sonata为您提供'sonata_type_choice_field_mask'类型，允许您根据此'sonata_type_choice_field_mask'输入的值动态更改表单上显示的字段。

Here is the doc您可以在其中找到有关奏鸣曲类型和选择字段掩码的所有内容。

protected function configureFormFields(FormMapper $formMapper)
{
    $formMapper
        ->add('reference', 'sonata_type_choice_field_mask', array(
            'choices' => array(
                //The list of available 'Reference' here
                'choice1',
                'choice2'
            ),
            'map' => array(
                //What you want to display depending of the selected option
                'choice1' => array(
                    // List of the fields displayed if choice 1 is selected
                    'field1', 'field3'
                ),
                'choice2' => array(
                    // List of the fields displayed if choice 2 is selected
                    'field2', 'field3'
                )
            ),
            'placeholder' => 'Choose an option',
            'required' => false
        ))
        ->add('field1', 'sonata_type_model_list', array(/* Options goes here */))
        ->add('field2', 'url', array(/* Options goes here */))
        ->add('field3')
    ;
}

Answer 2

您似乎可以访问直接使用 Sonata\AdminBundle\Admin\AbstractAdmin::getForm 构建表单的管理员。

$form = $this->admin->getForm();

使用Sonata字段类型在Controller中创建表单

2 个答案: