ajax回调在视图样式插件中工作吗？

Question

我的模块有一个自定义视图风格插件。在样式设置表单中，我希望实现ajax回调，如以下所述：https://www.drupal.org/docs/8/api/javascript-api/ajax-forms

然而，我没有任何运气。我可以得到旋转的效果，但之后什么都不会发生。下面是我尝试过的一个例子：

use Drupal\Core\Ajax\AjaxResponse;
use Drupal\Core\Ajax\HtmlCommand;  

  public function TestCallback(array $form, FormStateInterface $form_state) {
    $ajax_response = new AjaxResponse();
    $text = 'Text to see if this is working';
    $ajax_response->addCommand(new HtmlCommand('#foo', $text));

    return $ajax_response;
  }

  /**
   * {@inheritdoc}
   */
  public function buildOptionsForm(&$form, FormStateInterface $form_state) {
    parent::buildOptionsForm($form, $form_state);

    $form['test'] = [
      '#title' => t('TEST!'),
      '#type' => 'select',
      '#options' => [
        '' => t('None'),
        'foo' => t('Bar'),
      ],
      '#ajax' => [
        'event' => 'change',
        'callback' => '::TestCallback',
    // having the line below uncommented seems to help        
    //'url' => views_ui_build_form_url($form_state),
      ],
      '#prefix' => '',
      '#default_value' => '',
    ];
  }

如果取消注释“url”，将得到以下错误：

Uncaught TypeError: Cannot read property 'settings' of undefined
at resetSize (:34:32)
at later (debounce.js?v=8.6.3:20)

如果我继续评论的话，我会得到：

Failed to load resource: the server responded with a status of 500 () 
/admin/structure/views/ajax/display/view_name/page_1/style_options? 
_wrapper_format=drupal_ajax&ajax_form=1&_wrapper_format=drupal_ajax:1
Uncaught Drupal.AjaxError
ajax.js?v=8.6.3:500

缺少和/或不正确的是什么？

Answer 1

据我所知，看到AJAX文档，我认为您不应该返回一个AjaxResponse对象。您应该从Ajax回调返回一个可呈现的数组。

示例代码

public function buildOptionsForm(&$form, FormStateInterface $form_state) {
  parent::buildOptionsForm($form, $form_state);
  $form['test'] = [
    '#title' => t('TEST!'),
    '#type' => 'select',
    '#options' => [
      '' => t('None'),
      'foo' => t('Bar'),
    ],
    '#ajax' => [
      'event' => 'change',
      'wrapper' => 'ajax-output-container',
      // Since views style options forms are complex, they're built by
      // Drupal in a different way. To bypass this problem we need to
      // provide the full path to the Ajax callback.
      'callback' => __CLASS__ . '::testCallback',
    ],
    '#prefix' => '',
    '#default_value' => '',
  ];
}

/**
 * This is the AJAX callback.
 *
 * It generates the output for the AJAX request. The output REPLACES
 * the "wrapper" you define in #ajax.
 */
public function testCallback(array $form, FormStateInterface $form_state) {
  // This element will replace the DIV#foo you have in your #prefix.
  $elem = [
    '#type' => 'markup',
    '#markup' => 'Is it not working!>',
  ];
  return $elem;
}

调试

• 确保在页面加载时，字段前缀中有正确ID的div。您可以在前缀中添加一些默认文本，以便在页面加载时看到它？
• 如果由于某种原因，HTML标记被从#prefix标记中删除，那么尝试在表单中插入一个单独的项目，并使用#markup。

Answer 2

$route_name = "views_ui.form_{$form_key}"；可能是问题所在。

尝试在views_ui_build_form_url($form_state)->toString()；函数下面打印这个结果，看看是否是问题所在: buildOptionsForm

如果保留注释，则会发生错误，因为它们不是ajax所需的url或“#包装器”。

在这里，您可以更深入地研究这个错误：https://api.drupal.org/api/drupal/core%21modules%21views_ui%21 admin.inc/函数/视图_用户界面_构建_表格_url/8.2.x

Answer 3

我认为您可能会遇到一个与这个Drupal核心问题相关的问题。请参阅有关实现工作的发行评论和这一个。