Question

我正在参与Symfony项目，所以这就是我的问题：

我有一个实体，对于该实体，我在同一页面中有两个表单（一个用于插入，另一个用于更新插入的）。所以我做到了。

ActionMethod

public function adminTareasAction(Request $request) {

    $newTareaForm = $this->createForm(TareaType::class, null, array("formType" => "newTarea"));
    $editTareasForm = $this->createForm(TareaType::class, null, array("formType" => "editTareas"));

    if($request->isMethod("POST")) {
        if(!is_null($request->request->get('newTarea'))) {
            $newTareaForm->handleRequest($request);
            if($newTareaForm->isSubmitted() && $newTareaForm->isValid()) {
                $newTarea = $newTareaForm->getData();

                $dataManager = $this->getDoctrine()->getManager();
                $dataManager->persist($newTarea);
                $dataManager->flush();

                return $this->redirectToRoute("admin_tareas");
            }
        }
        elseif(!is_null($request->request->get('editTareas'))) {
            $editTareasForm->handleRequest($request);
            if($editTareasForm->isSubmitted() && $editTareasForm->isValid()) {
                $newTarea = $newTareaForm->getData();

                $dataManager = $this->getDoctrine()->getManager();
                $dataManager->persist($newTarea);
                $dataManager->flush();

                return $this->redirectToRoute("admin_tareas");
            }
        }
    }

    $tareas = $this->getDoctrine()->getRepository('FabricacionBundle:Tarea')->findAll();
    if(!$tareas) {
        $tareas = "No hay Tareas";
    }
    return $this->render('UsersBundle:Admin:adminTareas.html.twig', array("newTareaForm" => $newTareaForm->createView(), "editTareasForm" => $editTareasForm->createView(), "tareas" => $tareas));
}

类型类

class TareaType extends AbstractType {

    private $formType;

    /**
     * {@inheritdoc}
     */
    public function buildForm(FormBuilderInterface $builder, array $options) {
        $this->formType = $options["formType"];
        if($this->formType == "newTarea") {
            //var_dump();
            $builder
                ->add('tareaName', TextType::class)
                ->add('tareaOrden', IntegerType::class)
                ->add('submitNewTarea', SubmitType::class);
        }
        elseif($this->formType == "editTareas") {
            $builder
                ->add('newName', CollectionType::class, array("entry_type" => TextType::class, "allow_add" => true))
                ->add('newOrden', CollectionType::class, array("entry_type" => IntegerType::class, "allow_add" => true))
                ->add('deleteTarea', CollectionType::class, array("entry_type" => CheckboxType::class, "allow_add" => true))
                ->add('submitTarea', SubmitType::class);

        }
    }

    /**
     * {@inheritdoc}
     */
    public function configureOptions(OptionsResolver $resolver) {
        $resolver->setDefaults(array(
            'data_class' => 'FabricacionBundle\Entity\Tarea',
            "formType" => null
        ));
    }

    /**
     * {@inheritdoc}
     */
    public function getBlockPrefix() {
        /*if($this->formType == "newTarea") {
            return $this->formType;
        }
        elseif($this->formType == "editTareas") {
            return $this->formType;
        }*/
        return 'FabricacionBundle_tarea';
    }
}

我唯一需要的是更改表单名称，然后按名称在控制器中处理它们。

Answer 1

只需删除getBlockPrefix()。您正在通过其FQCN访问您的字段类型：

 $newTareaForm = $this->createForm(TareaType::class, null, array("formType" => "newTarea"));

所以你不需要为它设一个别名（不推荐使用）。

Answer 2

好吧，阅读Form类我得到了我必须做的事。

就这样创建表单：

$newTareaForm = $this->container->get('form.factory')->createNamedBuilder("newTarea", TareaType::class, null, array("formType" => "newTarea"))->getForm();
$editTareasForm = $this->container->get('form.factory')->createNamedBuilder("editTareas", TareaType::class, null, array("formType" => "editTareas"))->getForm();

Symfony buildForm构建器setName

2 个答案: