jquery - yii2 依赖的自动完成小部件

Question

我的表单中有两个字段:

<?= $form->field($model, 'bill_country')->widget(AutoComplete::className(), ['options' => ['class' => 'form-control'], 'clientOptions' => ['source' => $country_name]])->label('Country') ?>
<?= $form->field($model, 'bill_zip')->widget(AutoComplete::className(), ['options' => ['class' => 'form-control'], 'clientOptions' => ['source' => '/city/list/+id']])->label('Zip') ?>

根据第一个 block 的选择，我想更改第二个 block 的“源”。 +id 是一个 JavaScript 变量...因此，我想使用 val() 获取第一个自动完成的值，将其分配给 var id，然后将此 id 传递给第二个自动完成的源。我可以这样做吗？或者我应该使用其他选项？

谢谢!

Answer 1

所以，代码:

$this->registerJs("
    var country = '';
    $('#partner-bill_country').autocomplete({
        select: function( event, ui ) {
            country = (ui.item.country_id);
            $('#partner-bill_zip').autocomplete({
                source: '/city/list/'+country
            });
        }
    });
$('#partner-bill_zip').autocomplete({
    select: function(event, ui) {
        $('#partner-bill_city').val(ui.item.citname);
    }
});
", View::POS_READY, 'partner_script');
//initial arrays:
$city_zip = frontend\models\City::find()
    ->select(['CONCAT(cit_zip, " ", cit_name) as label', 'cit_zip as value', 'cit_id as id', 'cit_name as citname'])
    ->asArray()
    ->all();

$country_name = frontend\models\Country::find()
    ->select(['CONCAT(country_code, " ", country_name) as label', 'country_name as value','id as country_id'])
    ->asArray()
    ->all();
?>
<div class="partner-form">
    <?php $form = ActiveForm::begin(); ?>
        <div class="row">
            <div class="col-lg-6">
                  <?= $form->field($model, 'bill_country')->widget(AutoComplete::className(), ['options' => ['class' => 'form-control'], 'clientOptions' => ['source' => $country_name]])->label('Country') ?>
                  <?= $form->field($model, 'bill_zip')->widget(AutoComplete::className(), ['options' => ['class' => 'form-control'], 'clientOptions' => ['source' => $city_name]])->label('Zip') ?>