Symfony4：渲染无法传递参数

Question

从Symfony2.X更新到Symfony 4.4并验证运行后，发现Twig不识别模态
由于不识别modal，所以即使if语句中不以modal显示的代码以modal显示也会显示。
当我尝试“if modal is empty”时，显示的是if语句中的代码，所以好像modal没有通过
通过模式的代码随着 Symfony 的更新而发生了变化。这可能有影响。
你有什么想法吗？

Old_code

    <div id="imageDialog" title="Image management">
        {{ render(controller("AppBundle:Shop/Image:manager", {"modal": true})) }}
    </div>

Changed_code

    <div id="imageDialog" title="Image management">
        {{ render(controller("AppBundle\Controller\Shop\ImageController::managerAction", {"modal": true})) }}
    </div>

有问题的代码

    <div class="tabpanel selected" id="imageFolder" >
        <div class="tabcontent">
            <div class="operations">
                {% if not modal %}
                    <div class="pull-right">
                        <form method="post" action="{{ path('app_shop_image_delete') }}">
                            <input type="hidden" name="methods" value="delete">
                            <input type="hidden" name="_csrf_token" value="{{ csrf_token("authenticate") }}">
                            <input type="hidden" name="ids" value="[]">
                            <button type="submit" class="delete btn btn-danger"><i class="icon-trash"></i> Delete</button>
                        </form>
                    </div>
                {% endif %}
            </div>
        </div>
    </div>

图像控制器

     *
     * @Method("GET")
     * @Route("/manager")
     *
     * @Template("@AppBundle/Shop/Image/manager.html.twig")
     */
    public function managerAction(Request $request)
    {
        $routeParams = $request->get('_route_params');
        $uploadUrl = $this->generateUrl("app_shop_image_save");
        return array(
            'modal' => isset($routeParams["modal"]) ? $routeParams["modal"] : false,
            'form'  => $this->createUploadForm($uploadUrl)->createView(),
        );
    }

Answer 1

用下面的方法效果很好。我无法获取$routeParams["modal"]，但如果有办法获取，请告诉我。

    public function managerAction(Request $request)
    {
        $routeParams = $request->query->get('modal');
        $uploadUrl = $this->generateUrl("app_hq_image_save");
        return array(
            'modal' => isset($routeParams) ? $routeParams : false,
            'form'  => $this->createUploadForm($uploadUrl)->createView(),
        );
    }

     <div id="imageDialog" title="Image management">
         {{ render(controller("AppBundle\Controller\Shop\ImageController::managerAction", {}, {"modal": true})) }}
     </div>

Answer 2

public function managerAction(Request $request)
{
    $isModal = $request->query->get('modal') != null ? $request->query->get('modal') : false;
    $uploadUrl = $this->generateUrl("app_hq_image_save");
    return array(
        'modal' => $isModal,
        'form'  => $this->createUploadForm($uploadUrl)->createView(),
    );
}

但我想知道您是否不应该将模态作为具有默认值 false 的真实路由参数。 https://symfony.com/doc/4.4/routing.html#optional-parameters

Symfony4：渲染无法传递参数

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