Question

我对symfony2中的管理表单有疑问。所以在我的控制器中我有：

$form = $this->createForm(SubMenuType::class);

在SubMenuType类中我有：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $aTemplate  = Template::$aTemplates;
    $aMenu = $this->getDoctrine()->getRepository('AppAdminBundle:Menu')->findAll();
    $builder
        ->add('name', TextType::class)
        ->add('template',ChoiceType::class, array('choices' => $aTemplate,'choices_as_values' => true))
        ->add('menu',ChoiceType::class, array('choices' => $aMenu,'choices_as_values' => true))
        ->add('save',SubmitType::class, array('label'=> 'Save'))
    ;
}

问题是：如何从$aMenu中的Menu实体获取SubMenuType数据？请帮我。 Thx提前

Answer 1

解决问题的最简单方法是使用EntityType表单类型：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $aTemplate  = Template::$aTemplates;
    $builder
        ->add('name', TextType::class)
        ->add('template',ChoiceType::class, array('choices' => $aTemplate,'choices_as_values' => true))
        ->add('menu', EntityType::class, array(
            'class' => 'AppAdminBundle:Menu', 
            'choice_label' => 'property of your menu entity you want as label'
        ))
        ->add('save',SubmitType::class, array('label'=> 'Save'))
    ;
}

将数据从控制器发送到表单类型

1 个答案: