Question

我正在建立一个在网上报纸上写文章的后端管理员，我需要插入图片（1对n关系）。事实上，我写了一个表单来处理要显示的字段，这很好。

现在我想添加一个分部，它将显示已经链接到文章的图像，所以我有一个组件类，其中包含一个操作executeShowImages和一个模板_showImages.php。

My question I would to keep all the definition in the form class, but just add this component/partial display and I cannot do it easily. Here is my generator file:

    generator:
  class: sfPropelGenerator
  param:
    model_class:           article
    theme:                 admin
    non_verbose_templates: true
    with_show:             false
    singular:              article
    plural:                articles
    route_prefix:          article
    with_propel_route:     1
    actions_base_class:    sfActions

    config:
      actions: ~
      fields:  ~
      list:    ~
      filter:  ~
      form:
        class: ArticleFormBackend
        display: [ ~showImages ]
      edit:    ~
      new:     ~

如果以这种方式执行，则只显示showImage组件。怎么能做到？

非常感谢你的帮助

Answer 1

Yes you are right - only showImage component is displayed . You should to insert list of all fields you want to see (if you override option).

所以你的表单选项应该是这样的：

  form:
    class: ArticleFormBackend
    display: [ title, body, ~showImages ]

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