[英]Yii2 - Show form fields based on dropDownList
I have the following dropDownList 我有以下dropDownList
<?= $form->field($model, 'moradaalternativa')
->dropDownList(
[
'Não' => 'Não',
'Sim' => 'Sim'],
['prompt'=>'Faça a sua escolha'],
);
?>
What i'm trying to do is: If the value is = Sim, then some other form fields that are hidden by a css class show up below the dropDownList, and if the value is = Não, then the form fields hide from the page again. 我想做的是:如果值是= Sim,则由CSS类隐藏的其他一些表单字段会显示在dropDownList下方,如果值是=Não,则表单字段将从页面中隐藏再次。
I know that there is a 'onchange' property like javascript but i don't know how to apply it to this effect or even if i need to use it. 我知道有一个像javascript这样的'onchange'属性,但是我不知道如何将其应用于这种效果,甚至我都不需要使用它。
Any ideas? 有任何想法吗?
It's pretty simple, all you need is: 这很简单,您需要做的是:
$(document).ready(function () {
$(document.body).on('change', '#your-id', function () {
var val = $('#your-id').val();
if(val > 0 ) {
$('.class').hide();
} else {
$('.class').show();
}
});
});
And just change the names as needed. 并根据需要更改名称。 For Yii2 you can wrap it, then you can just put the code in a view file but it's better to put it in a JS file:
对于Yii2,您可以将其包装,然后将代码放入视图文件中,但最好将其放入JS文件中:
<?php
$script = <<< JS
code here
JS;
$this->registerJs($script);
?>
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