Question

我是symfony2的新手，我必须从3个表/对象创建一个选择表单：

**Game table:**
id,another_column
game_1, 2
game_2, 4
game_3, 10
game_4, 1

**Score table:**
id,user_id,game_id
1,4,game_1
2,4,game_3

用户通过身份验证后（我使用的是软件用户捆绑包），我必须创建一个包含所有未播放游戏的选择表单。在这种情况下，我的选择表单需要有两个选项（game_2和game_4）。

ScoreFormType.php

<?php
/**
 * @package evaluation
 */
namespace GameBundle\Form\Type;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;

class ScoreFormType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder->add('game');
    }

    public function getName()
    {
        return 'game_score';
    }  
}

DefaultController.php

<?php

namespace AppBundle\Controller;

use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\ORM\EntityRepository;
use GameBundle\Entity\Score;
use GameBundle\Form\Type\ScoreFormType;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;

class DefaultController extends Controller
{
    /**
     * @return \Symfony\Component\HttpFoundation\Response
     */
    public function indexAction(Request $request)
    {    
        $m = $this->getDoctrine()->getManager();
        $parameters = [];
        if (null !== $this->getUser()) {
            $score = new Score();
            $score
                ->setUser($this->getUser())
                ->setPoints(rand(1,50))
            ;
            $form = $this->createForm(new ScoreFormType(), $score);

            $parameters['form'] = $form->createView();
        }

        return $this->render('AppBundle::index.html.twig', $parameters);
    }

}

有没有可以帮助我的例子？我试图做一项研究，但没有任何相关性。

谢谢。

Answer 1

您应首先从数据库中获取游戏，然后将其添加到表单中：

$form = $this->createForm(new ScoreFormType(), $games);

在ScoreFormType.php中创建：

public function __construct($aYourGames = array())
{
    $this->aYourgames = $aYourGames['Your data'];
}

现在创建“选择”：

->add('games', 'choice', array(
            'required' => true,
            'label' => 'games',
            'choices' => $this->aYourgames
            )
        )

您还可以为您的选择提供课程，当您使用Javascript时，它会有所帮助：

'attr' => array('class' => 'govChoice'),

Answer 2

我假设以下实体：

游戏：

class Game 
{
    private $id;
    private $title;
}

<强>得分：

class Score
{
    private $id;
    private $user_id; 

    /** 
     * @ORM\ManyToOne(targetEntity="Game")
     */
    private $game_id;
}

<强> FormType ：

<?php
/**
 * @package evaluation
 */
namespace GameBundle\Form\Type;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;

class ScoreFormType extends AbstractType
{

    private $user;

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

    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $user = $this->user;

        $builder->add('game', 'entity', array(
                        'class' => 'GameBundle:Game',
                        'property' => 'title',
                        'query_builder' => function (EntityRepository $er) use ($user) {
                               return $er->createQueryBuilder('game')
                                    ->where('game.id NOT IN(SELECT score.game_id FROM GameBundle:Score score WHERE score.user_id = :user'))
                                    ->setParameter('user', $user)
                                    ->orderBy('game.title', 'ASC');
                     ));
    }

    public function getName()
    {
        return 'game_score';
    }  
}

<强> DefaultController ：

<?php

namespace AppBundle\Controller;

use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\ORM\EntityRepository;
use GameBundle\Entity\Score;
use GameBundle\Form\Type\ScoreFormType;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;

class DefaultController extends Controller
{
    /**
     * @return \Symfony\Component\HttpFoundation\Response
     */
    public function indexAction(Request $request)
    {    
        $m = $this->getDoctrine()->getManager();
        $parameters = [];
        $form = $this->createForm(new ScoreFormType($this->getUser()));

        $parameters['form'] = $form->createView();

        return $this->render('AppBundle::index.html.twig', $parameters);
    }

}

也许您需要在表单中稍微调整查询，但这可能是使用表单组件提供的基本工具解决它的一种可能性。

symfony2选择具有实体关系的表单

2 个答案: