Question

我有一个包含1个EntityType字段的表单，该字段必须包含根据未在第一个实体中映射的第二个EntityType字段的选项，如下所示：

ServicePlaceType.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只显示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，但我不知道如何在formType中获取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？或者如何通过其他方式动态过滤选项？

感谢您的帮助！

Answer 1

您可以使用jquery库更简单一点：

首先，我们更改构建器，使用<option data-type="...">选项将地点类型ID呈现为choice_attr：

$builder
    ->add('placetype', EntityType::class, array(
        "class" => "AppBundle:PlaceType",
        "mapped" => false
    ))
    ->add('idplace', EntityType::class, array(
        "class" => "AppBundle:Place",
        'choice_attr' => function ($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>

Symfony - FormType - 动态选择

1 个答案: