6

我不知道如何为 symfony2 中的实体类型设置默认值。我的代码如下所示:

$rewardChoice = $this->createFormBuilder($reward)
    ->add('reward_name', 'entity', array(
        'class' => 'FuelFormBundle:Reward',
        'property' => 'reward_name',
        'data' => 2,
        'query_builder' => function(EntityRepository $er){
            return $er->createQueryBuilder('r')
                ->where('r.active = 1')
                ->groupBy('r.reward_id')
                ->orderBy('r.reward_name', 'DESC');

        },
    ))
    ->getForm();

但是,您需要交出您正在使用的对象才能使其工作。我的答案如下。

我在这方面找到了很多不同的答案,但它们都重组了表单的构建方式。这要容易得多。

4

4 回答 4

9

所以我找到了很多答案来完成这项工作,但他们似乎都以另一种方式重组了要构建的表单,但是我发现设置对象效果最好,所以我想我会发布我的解决方案,以防有人遇到问题再次。

这是我在控制器中的代码。

// this is setting up a controller and really isn't important 
$dbController = $this->get('database_controller');

// this is getting the user id based on the hash passed by the url from the
// database controller
$user_id = $dbController->getUserIdByHash($hash);

// this is getting the Reward Entity.  A lot of times you will see it written as 
// $reward = new Reward however I am setting info into reward right away in this case
$reward = $dbController->getRewardByUserId($user_id);


$rewardChoice = $this->createFormBuilder($reward)
    ->add('reward_name', 'entity', array(
        'class' => 'FuelFormBundle:Reward',
        'property' => 'reward_name',

        // I pass $reward to data to set the default data.  Whatever you
        // assign to $reward will set the default value.  
        'data' => $reward,  
        'query_builder' => function(EntityRepository $er){
                return $er->createQueryBuilder('r')
                    ->where('r.active = 1')
                    ->groupBy('r.reward_id')
                    ->orderBy('r.reward_name', 'DESC');
        },
    ))
    ->getForm();

我希望这能让事情更清楚。我看到了很多相同的问题,但没有这个解决方案。

于 2013-10-24T22:34:13.530 回答
6

最好的方法是在创建表单之前设置奖励名称。例如:

$reward->setRewardName('your_relationship_reference_here');

$rewardChoice = $this->createFormBuilder($reward)

使用该data字段可能会导致问题。

于 2018-03-26T12:07:39.333 回答
2

对于 symfony 4,我们可以使用 'placeholder' => 'Choose an option',

->add('reward_name', EntityType::class, array(
                    'class' => Reward::class,
                    'choice_label' => 'name',
                    'placeholder' => 'Choose an option',
                ))
于 2018-08-24T05:50:32.777 回答
1

从表单的目标对象初始化值

$article=2/*Id of item that you wont to set as default*/    
$stock->setArtigo($this
                ->getDoctrine()
                ->getManager()
                ->getRepository('AppBundle:Artigo')
                ->find($article));

            $form = $this->createForm('AppBundle\Form\StockType', $stock);
于 2018-08-09T19:49:59.923 回答