symfony - 如何在 EasyAdmin 3 中添加自定义操作？

Question

我的实体 Participant 有一个 CrudController。我想添加一个自定义操作 sendAcknowledgementEmail。 EasyAdmin docs没有提及有关自定义函数参数或返回值的任何内容。
我有以下代码

public function configureActions(Actions $actions): Actions
{
    $send_acknowledgement_email = Action::new('sendAcknowledgementEmail', 'Send Acknowledgement Email', 'fa fa-send')
        ->linkToCrudAction('sendAcknowledgementEmail');

    return $actions
        ->add(Crud::PAGE_INDEX, $send_acknowledgement_email)
        ->add(Crud::PAGE_EDIT, $send_acknowledgement_email)
    ;
}

public function sendAcknowledgementEmail() //Do I need parameters?
{
    //How do I get the Entity?

    //What should I return?
}

到目前为止，EasyAdmin 检测到自定义函数，但我收到错误消息“ Controller 必须返回一个“Symfony\Component\HttpFoundation\Response”对象，但它返回了 null。您是否忘记在 Controller 中的某处添加 return 语句？
我如何从这里继续？

Answer 1

捆绑包的 v3.x 是相当新的，文档还不完善。
基于 Ceochronos 的回答，这是我对克隆操作的实现。

public function configureActions(Actions $actions): Actions
{
    $cloneAction = Action::new('Clone', '')
        ->setIcon('fas fa-clone')
        ->linkToCrudAction('cloneAction');

    return $actions
        ->add(Crud::PAGE_INDEX, $cloneAction);

}

public function cloneAction(AdminContext $context)
{
    $id     = $context->getRequest()->query->get('entityId');
    $entity = $this->getDoctrine()->getRepository(Product::class)->find($id);

    $clone = clone $entity;

    // custom logic 
    $clone->setEnabled(false);
    // ...
    $now = new DateTime();
    $clone->setCreatedAt($now);
    $clone->setUpdatedAt($now);

    $this->persistEntity($this->get('doctrine')->getManagerForClass($context->getEntity()->getFqcn()), $clone);
    $this->addFlash('success', 'Product duplicated');

    return $this->redirect($this->get(CrudUrlGenerator::class)->build()->setAction(Action::INDEX)->generateUrl());
}