symfony2选择具有实体关系的表单 - symfony2 select form with Entity Relations

我是symfony2的新手，我必须从3个表/对象中创建一个选择表单：

**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

在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);
    }
}

有什么例子可以帮助我吗？我试着做了一项研究，但没有什么相关的。

谢谢。

您应该首先从数据库中获取游戏，然后使用将其添加到表单中

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

在ScoreFormType.php中创建以下内容：

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

现在创建您的"选择"：

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

你也可以给你的选择一个类，当你使用Javascript:时它会有所帮助

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

我假设以下实体：

游戏：

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

分数：

class Score
{
    private $id;
    private $user_id; 
    /** 
     * @ORM'ManyToOne(targetEntity="Game")
     */
    private $game_id;
}

表单类型：

<?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';
    }  
}

默认控制器：

<?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);
    }
}

也许您需要稍微处理一下表单中的查询，但这是使用表单组件提供的基本工具解决问题的一种可能性。