小编Udi*_*wat的帖子

当用户点击前端的链接时,我想在弹出窗口中显示我的自定义表单,我尝试了很多在网络上显示的解决方案,但对我不起作用.

这是我的代码.

        $response = new AjaxResponse();
        // Get the modal form using the form builder.
        $modal_form = $this->formBuilder->getForm('Drupal\fwsactions\Forms\FwsActionsForm');
        $modal_form['#attached']['library'][] = 'core/drupal.dialog.ajax';
        // Add an AJAX command to open a modal dialog with the form as the content.
        $modal_form = render($modal_form);
        $response->addCommand(new OpenModalDialogCommand('My Modal Form', $modal_form, ['width' => '800']));

        return $response;

如何通过单击链接打开此表单应该怎么做.