Symfony2 -从数据库中添加选项


symfony2 - adding choices from database

我希望用自定义查询的值填充symfony2中的选择框。我已经尽可能地简化了。

控制器

class PageController extends Controller
{
    public function indexAction()
    {
      $fields = $this->get('fields');
      $countries =  $fields->getCountries(); // returns a array of countries e.g. array('UK', 'France', 'etc')
      $routeSetup = new RouteSetup(); // this is the entity
      $routeSetup->setCountries($countries); // sets the array of countries
      $chooseRouteForm = $this->createForm(new ChooseRouteForm(), $routeSetup);

      return $this->render('ExampleBundle:Page:index.html.twig', array(
        'form' => $chooseRouteForm->createView()
      ));
    }
}

ChooseRouteForm

class ChooseRouteForm extends AbstractType
{
  public function buildForm(FormBuilderInterface $builder, array $options)
  {
    // errors... ideally I want this to fetch the items from the $routeSetup object 
    $builder->add('countries', 'choice', array(
      'choices' => $this->routeSetup->getCountries()
    ));
  }
  public function getName()
  {
    return 'choose_route';
  }
}

您可以使用..将选择传递给表单

$chooseRouteForm = $this->createForm(new ChooseRouteForm($routeSetup), $routeSetup);

那么在你的表单中…

private $countries;
public function __construct(RouteSetup $routeSetup)
{
    $this->countries = $routeSetup->getCountries();
}
public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder->add('countries', 'choice', array(
        'choices' => $this->countries,
    ));
}

更新(和改进)2.8 +

首先,你真的不需要传入国家作为路由对象的一部分,除非它们将被存储在DB中。

如果在DB中存储可用的国家,则可以使用事件侦听器。如果没有(或者不想使用侦听器),可以在选项区域中添加国家。

利用期权

In the controller.

$chooseRouteForm = $this->createForm(
    ChooseRouteForm::class,
    // Or the full class name if using < php 5.5
    $routeSetup,
    array('countries' => $fields->getCountries())
);

在你的形式…

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder->add('countries', 'choice', array(
        'choices' => $options['countries'],
    ));
}
public function configureOptions(OptionsResolver $resolver)
{
    $resolver
        ->setDefault('countries', null)
        ->setRequired('countries')
        ->setAllowedTypes('countries', array('array'))
    ;
}

使用侦听器(如果模型中有国家数组)

In the controller.

$chooseRouteForm = $this->createForm(
    ChooseRouteForm::class,
    // Or the full class name if using < php 5.5
    $routeSetup
);

在你的形式…

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->addEventListener(FormEvents::PRE_SET_DATA, function(FormEvent $event) {
            $form = $event->getForm();
            /** @var RouteSetup $routeSetup */
            $routeSetup = $event->getData();
            if (null === $routeSetup) {
                throw new 'Exception('RouteSetup must be injected into form');
            }
            $form
                ->add('countries', 'choice', array(
                    'choices' => $routeSetup->getCountries(),
                ))
            ;
        })
    ;
}

我还不能评论或投票,所以我只能在这里回复Qoop的回答:除非您开始使用表单类型类作为服务,否则您的建议将有效。一般应该避免通过构造函数向表单类型对象添加数据。

表单类型类想象成——它是对表单的一种描述。当你创建一个实例的形式(通过构建它),你得到的对象的形式,是建立在表单类型的描述,然后填入数据。

看看这个:http://www.youtube.com/watch?v=JAX13g5orwo -这种情况在演示文稿的31分钟左右被描述。

你应该使用表单事件FormEvents::PRE_SET_DATA并在表单被注入数据时操作字段。见:http://symfony.com/doc/current/cookbook/form/dynamic_form_modification.html customizing-your-form-based-on-the-underlying-data

我得到它的工作通过调用getData上的生成器

FormBuilderInterface $builder

//控制器
$myCountries = $this->myRepository->all(['continent' => 'Africa']);
$form = $this->createForm(CountriesType::class, $myCountries);
//FormType

use Symfony'Component'Form'Extension'Core'Type'ChoiceType;
public function buildForm(FormBuilderInterface $builder, array $options): void
    {
        $builder
            ->add('pages', ChoiceType::class, [
                'choices' => $builder->getData()
            ])
        ;
    }