javascript - Symfony - FormType - 动态选择

标签 javascript php symfony

我有一个带有 1 个 EntityType 字段的表单,该字段必须包含取决于第二个 EntityType 字段的选项,该字段未映射到第一个实体中,如下所示:

服务类型.php :

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('placetype', EntityType::class, array(
            "class" => "AppBundle:PlaceType",
            "choice_label" => "place",
            "mapped" => false
        ))
        ->add('idplace', EntityType::class, array(
            "class" => "AppBundle:Place",
            "choice_label" => "place"
        ))
        ->add('...');

表格

+---------+--------------+---------------+-----------+
| Service | ServicePlace |     Place     | PlaceType |
+---------+--------------+---------------+-----------+
|         | id           |               |           |
+---------+--------------+---------------+-----------+
|         | idplace >    | < id          |           |
+---------+--------------+---------------+-----------+
| id >    | < idservice  | idPlaceType > | < id      |
+---------+--------------+---------------+-----------+
| service |              | place         | placetype |
+---------+--------------+---------------+-----------+

因此,当我选择一个 PlaceType 时,我希望 Place select 仅显示 idplacetype 与 PlaceType id 匹配的地方。

我尝试在 javascript 中使用 PlaceType 选择上的 onChange 事件,根据 PlaceType 实际值过滤 Place 选项,但我不知道如何获取 formType 中 Place 的 PlaceType 属性。 我试过那种东西,但它不起作用

->add('idplace', EntityType::class, array(
            "class" => "AppBundle:Place",
            "choice_label" => "place",
            "attr" => array("placeType" => $this->getPlaceType()), // nor like that
  ))

  ->add('idplace', EntityType::class, array(
            "class" => "AppBundle:Place",
            "choice_label" => "place",
            "attr" => array("placeType" => function ($place) {
                return $place->getPlaceType();
            }), // neither like that
   ))

有人知道如何获取这些数据吗?或者如何通过另一种方式动态过滤选项?

感谢您的帮助!

最佳答案

您可以使用 jquery 来完成图书馆更简单一点:

首先,我们稍微更改构建器以将地点类型 id 渲染为 <option data-type="...">使用 choice_attr选项:

$builder
    ->add('placetype', EntityType::class, array(
        "class" => "AppBundle:PlaceType",
        "mapped" => false
    ))
    ->add('idplace', EntityType::class, array(
        "class" => "AppBundle:Place",
        'choice_attr' => function ($place) {
            // output: <option data-type="...">...</option>
            return array('data-type' => $place->getPlaceType()->getId());
        },
    ))

接下来,在你的模板中:

{# ... #}

{{ form(form) }}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.2.4/jquery.min.js"></script>
<script>
    // when the 1st <select> was changed, then update the 2nd 
    // from current value and data-type option attribute.
    $(document).on('change', '#form_placetype', function () {
        var $idplace = $('#form_idplace'),
            // current value
            placetype = $(this).val(),
            // select available options from current value
            $available = $idplace.find('option[data-type="' + placetype + '"]');

        // deselect when the 1st <select> has changed.
        $idplace.val('');
        // hide no available options from current value
        $idplace.find('option').not($available).hide();
        // show available options from current value
        $available.show();
    });

    // Update 2nd <select> on page load.
    $('#form_placetype').trigger('change');
</script>

关于javascript - Symfony - FormType - 动态选择,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39400337/

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