构建表单与单表继承 [英] Build Form with Single Table Inheritance

问题描述

我有我的实体文章和一个单独的表继承如下：

I have my entity Article and one single table inheritance like this :

/**
 * Article
 *
 * @ORM\Table(name="article")
 * @ORM\Entity(repositoryClass="PM\PlatformBundle\Repository\ArticleRepository")
 * @ORM\InheritanceType("SINGLE_TABLE")
 * @ORM\DiscriminatorColumn(name="media", type="string")
 * @ORM\DiscriminatorMap({"article" = "Article", "movie" = "Movie", "image" = "Image", "text" = "Text"})
 */
class Article
{
    protected $id;
    protected $title;
    protected $description;
    protected $author;
    //other attributes and setters getters
}

class Image extends Article
{
    private $path;
    //getter setter
}

 class Movie extends Article
{
    private $url;
    //getter setter
}

所以我的文章的对象类型是Image或电影或文字。好的，现在我想建立一个表单，用户可以发布一个新的文章：在这种形式，用户必须选择树型（3无线电按钮）：图像或电影或文本，当然还有其他领域：标题和描述。我该怎么办？因为使用命令

So my article's object type is either Image or movie or text only. Ok now I would like build a form wherein users can post a new article : in this form, the user has to choice between tree type (3 radios button) : image OR movie OR text only and of course the other fields : title and description. How I can do that ? Because with the command

php bin / console doctrine：generate：form myBundle：Article

php bin/console doctrine:generate:form myBundle:Article

呈现的表单为：

class ArticleType extends AbstractType
{
    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('title',          TextType::class)
            ->add('description',    TextareaType::class)
            ->add('save',           SubmitType::class);
        ;
    }

    /**
     * @param OptionsResolver $resolver
     */
    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'PM\PlatformBundle\Entity\Article'
        ));
    }
}

我不知道实施我的STI的方式这种形式的关系。因为我没有在我的文章实体/对象中的字段（仅在我的表中）。
我必须添加一个Custom ChoiceType（）字段，但它需要一个属性。
当我尝试以下列形式添加它：

I don't know the way to implement my STI relation in this form. Because I have not field in my Article entity/object for the type (only in my table). I have to add a Custom ChoiceType() field but it require a attribute. When I try to add this in the form :

        ->add('path',           SearchType::class)
        ->add('url',            UrlType::class)

我收到这个错误：

属性path也不是方法getPath（）之一，path（），isPath（），hasPath（），__get（）存在，并在类PM\PlatformBundle\Entity\Article中具有公共访问权限。

Neither the property "path" nor one of the methods "getPath()", "path()", "isPath()", "hasPath()", "__get()" exist and have public access in class "PM\PlatformBundle\Entity\Article".

因为我创建了一个Article的实例，而不是Image或Movie的实例。最初我创造了一个STI思维文章的新例子，也允许我定义文章的类型。但不是 ？

Because I have create an instance of Article, not an instance of Image or Movie. Initially I created a STI thinking a new instance of Article would allow me also to define the "type" of article. But not ? Right ?

推荐答案

您将需要制作三种表格（一种用于文章，一个用于 Movie ，另一个用于 Image ）。然后，在你的控制器中你必须选择处理它们：

You will have to make three forms (one for an Article, one for a Movie and one for an Image). Then, in your controller, you have to options to deal with them:

• 你可以使用一个动作来处理三种形式通过使用 $ form-> isSubmitted（））


• 通过表单创建一个操作，并且将每个表单的表单操作网址设置为正确的控制器。

• Either you use one action to handle the three forms (you can check wich one is submitted by using $form->isSubmitted())
• You create one action by form, and set the form action URL for each form to the correct controller.

最后，在您的模板中，将表单封装在div中，并使用我上一篇文章中的示例。

Finally, in your template, you encapsulate your forms in a div, and use the example in my previous post.

{% extends "CoreBundle::layout.html.twig" %}

{% block title %}{{ parent() }}{% endblock %}

{% block btn_scrollspy %}
{% endblock %}


{% block bundle_body %}
    <div class="well">
        <div class="selector">
            <input type="radio" name="form-selector" value="article-form"> Article
            <input type="radio" name="form-selector" value="movie-form"> Movie
            <input type="radio" name="form-selector" value="image-form"> Image
        </div>

        <div class="form article-form" style="display: none;">
            {{ form(articleForm) }}
        </div>
        <div class="form movie-form" style="display: none;">
            {{ form(movieForm) }}
        </div>
        <div class="form image-form" style="display: none;">
            {{ form(imageForm) }}
        </div>
    </div>
{% endblock %}

这篇关于构建表单与单表继承的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！