Question

我想知道是否有一种方法可以以不同于select的symfony形式“管理”多对一关联（例如，使用jQuery自动完成输入）。

我找到了，所以我阅读了有关this question

的文档

这是我的变压器类

// src/AppBundle/Form/DataTransformer/ZipcodesTransformer.php
namespace AppBundle\Form\DataTransformer;

use AppBundle\Entity\Issue;
use Doctrine\Common\Persistence\ObjectManager;
use Symfony\Component\Form\DataTransformerInterface;
use Symfony\Component\Form\Exception\TransformationFailedException;

class ZipcodesTransformer implements DataTransformerInterface {
    private $manager;
    public function __construct(ObjectManager $manager) {
        $this->manager = $manager;
    }

    public function transform($zipcode) {
        if (null === $zipcode) {
            return '';
        }
        return $zipcode->getId();
    }

    public function reverseTransform($zipcodeId) {
        if (!$zipcodeId) {
            return;
        }

        $zipcode = ->this->manager->getRepository('AppBundle:Zipcodes')->find($zipcodeId)
        ;

        if (null === $zipcode) {
            throw new TransformationFailedException(sprintf(
                'An zipcode with number "%s" does not exist!',
                $zipcodeId
            ));
        }

        return $zipcode;
    }
}

然后选择器Type class

// src/AppBundle/Form/ZipcodeSelectorType.php
namespace AppBundle\Form;

use AppBundle\Form\DataTransformer\ZipcodesTransformer;
use Doctrine\Common\Persistence\ObjectManager;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;
use Symfony\Component\Form\Extension\Core\Type\TextType;

class ZipcodeType extends AbstractType
{
    private $manager;

    public function __construct(ObjectManager $manager)
    {
        $this->manager = $manager;
    }

    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $transformer = new ZipcodesTransformer($this->manager);
        $builder->addModelTransformer($transformer);
    }

    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults(array(
            'invalid_message' => 'The selected ZIPCODE does not exist',
        ));
    }

    public function getParent()
    {
        return TextType::class;
    }
}

对象形式的类是

namespace AppBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\Extension\Core\Type\EmailType;
use Symfony\Component\Form\Extension\Core\Type\BirthdayType;
use Symfony\Component\Form\Extension\Core\Type\CheckboxType;
use AppBundle\Form\DataTransformer\ZipcodesTransformer;

class CustomerType extends AbstractType
{

    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('name', TextType::class ,array('label'=>'Customer Name'))
            ...
            ->add('email', EmailType::class)
            ->add('flag', CheckboxType::class, array('required' => false))
            ->add('zipcode', ZipcodeType::class)
        ;
    }

    /**
     * @param OptionsResolver $resolver
     */
    public function configureOptions(OptionsResolver $resolver) {
        $resolver->setDefaults(array('data_class' => 'AppBundle\Entity\Customer'));
    }
}

感谢Kern，代码现在正确且有效。现在为了“激活”已经测试过的jQuery自动完成，我修改了 ZipcodeSelectorType.php 并添加了这个函数

public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults(
        array (
            'invalid_message' => 'The selected ZIPCODE does not exist',
            'attr' => array('class'=>'zipcodeac'),//ADDED THIS LINE
            'placeholder'=>'Type to select a zipcode'//ADDED THIS LINE
          )
        );
    }

但是当渲染表单时，输入具有此类

class =“zipcodeac form-control ui-autocomplete-input”

和属性

自动填充= “关闭”

并且没有解雇Ajax。

Answer 1

您错过了use课程中使用ZipcodeType方法调用的getParent()语句。

添加解决当前错误的use Symfony\Component\Form\Extension\Core\Type\TextType;。

无论如何，如果你有超过数千＆＃34;您评论中描述的选项，更好的方法是根据您的鉴别器字段制作自动完成的网络服务:)！

Symfony形式的多个教义

1 个答案: