Question

我创建了一个基于实体字段的表单。此表单由允许用户选择实体的复选框组成（此处为＆＃39; car＆＃39;）。这工作正常但我需要自定义渲染以获得额外的信息。目前，显示的唯一信息是＆＃39; id＆＃39;属性。在我的情况下，我想以颜色，品牌等形式显示额外的实体信息...... 有谁知道怎么办？

控制器：

        public function chooserAction() {
        //symfony.com/doc/current/reference/forms/types/entity.html
        $cars = $this->getDoctrine()
                ->getRepository('CarBundle:Car')
                ->find(1);
        $formBuilder = $this->createFormBuilder();
        /*
          foreach ($cars as $car) {
          $formBuilder->add($car->getId() ,'checkbox')->getForm();
          }
         */
        $formBuilder->add('cars', 'entity', array(
            'class' => 'CarBundle:Car',
            'property' => 'id',
            'expanded' => 'true',
            'multiple' => 'true',
        ));

        $formBuilder->add('save', 'submit');
        $form = $formBuilder->getForm();
        $request = $this->get('request');
        $form->handleRequest($request);
        if ($form->isValid()) {
            echo "ok";

//            return $this->redirect($this->generateUrl('car_show', array('id' => $car->getId())));
        }

        return $this->render('CarBundle:Car:chooser.html.twig', array('form' => $form->createView()));
    }

实体：

<?php

namespace Foobar\CarBundle\Entity;

use Doctrine\ORM\Mapping as ORM;

/**
 * Car
 *
 * @ORM\Table()
 * @ORM\Entity(repositoryClass="Foobar\CarBundle\Entity\CarRepository")
 */
class Car
{
    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @var string
     *
     * @ORM\Column(name="name", type="string", length=255)
     */
    private $name;

    /**
     * @var string
     *
     * @ORM\Column(name="brand", type="string", length=255)
     */
    private $brand;

    /**
     * @var string
     *
     * @ORM\Column(name="color", type="string", length=255)
     */
    private $color;

    /**
     * @var integer
     *
     * @ORM\Column(name="power", type="integer")
     */
    private $power;


    /**
     * Get id
     *
     * @return integer 
     */
    public function getId()
    {
        return $this->id;
    }

    /**
     * Set name
     *
     * @param string $name
     * @return Car
     */
    public function setName($name)
    {
        $this->name = $name;

        return $this;
    }

    /**
     * Get name
     *
     * @return string 
     */
    public function getName()
    {
        return $this->name;
    }

    /**
     * Set brand
     *
     * @param string $brand
     * @return Car
     */
    public function setBrand($brand)
    {
        $this->brand = $brand;

        return $this;
    }

    /**
     * Get brand
     *
     * @return string 
     */
    public function getBrand()
    {
        return $this->brand;
    }

    /**
     * Set color
     *
     * @param string $color
     * @return Car
     */
    public function setColor($color)
    {
        $this->color = $color;

        return $this;
    }

    /**
     * Get color
     *
     * @return string 
     */
    public function getColor()
    {
        return $this->color;
    }

    /**
     * Set power
     *
     * @param integer $power
     * @return Car
     */
    public function setPower($power)
    {
        $this->power = $power;

        return $this;
    }

    /**
     * Get power
     *
     * @return integer 
     */
    public function getPower()
    {
        return $this->power;
    }
}

观点：

car chooser
{{ form(form) }}

Answer 1

您可以在Car实体中创建一个toString（）方法。

public function __toString()
{
    return '' . $this->getName();
}

如果你想要更多的想法，你可以遵循http://symfony.com/doc/current/cookbook/form/form_customization.html

如何在symfony 2中基于实体字段自定义表单渲染？

1 个答案: