4
Well I need to check the state of mine radio button
to make sure that if he is active in "Yes", he opens the <div class="camposExtras">
, click on "Not" he closes the div
, and this should be accessible at all times.
Mine radio button
down below:
<input type="radio" class="FlgPontua" name="FlgPontua" value="Sim" checked>
<input type="radio" class="FlgPontua" name="FlgPontua" value="Nao">
And mine div
is:
<div class="camposExtras">
Aqui vem os dados que é para esconder ou aparecer
</div>
I’m trying to do it like this:
$(".FlgPontua").change(function() {
console.log('entro aqui')
if ($(this).val() == "Sim") {
$('.camposExtras').show();
}else{
$('.camposExtras).hide();
}
});
But you’re not giving me anything back.
For some reason here it’s not working
– Renan Rodrigues
It here at Stackoverflow is working correctly. You have expressed the click on the button
Executar
code to see it in action?– Chun
Yes, I forgot an important detail using a different css icheck button
– Renan Rodrigues
the website is this http://icheck.fronteed.com/
– Renan Rodrigues
I got, thanks for the support
– Renan Rodrigues