php - 如何防止Symfony 4中出现ParseError？

Question

src/Controller/DataTableController.php

<?php

use DataTables\DataTablesInterface;

/**
 * Symfony 3.4 and above
 *
 * @Route("/users", name="users")
 *
 * @param Request $request
 * @param DataTablesInterface $datatables
 * @return JsonResponse
 */
public function usersAction(Request $request, DataTablesInterface $datatables): JsonResponse
{
    try {
        // Tell the DataTables service to process the request,
        // specifying ID of the required handler.
        $results = $datatables->handle($request, 'users');

        return $this->json($results);
    }
    catch (HttpException $e) {
        // In fact the line below returns 400 HTTP status code.
        // The message contains the error description.
        return $this->json($e->getMessage(), $e->getStatusCode());
    }
}

/**
 * Symfony 3.3 and below
 *
 * @Route("/users", name="users")
 *
 * @param Request $request
 * @return JsonResponse
 */
public function usersAction(Request $request): JsonResponse
{
    try {
        /** @var \DataTables\DataTablesInterface $datatables */
        $datatables = $this->get('datatables');

        // Tell the DataTables service to process the request,
        // specifying ID of the required handler.
        $results = $datatables->handle($request, 'users');

        return $this->json($results);
    }
    catch (HttpException $e) {
        // In fact the line below returns 400 HTTP status code.
        // The message contains the error description.
        return $this->json($e->getMessage(), $e->getStatusCode());
    }
}

我收到错误消息:

Answer 1

类的结构(与Symfony不相关)是

class MyClassController() {

  public function myMethodAction() {
  }
}

无法避免解析错误，因为这只是PHP的无效语法。