jquery simple hide show based on checkbox checked
Question
I want to show and hide the options inside the main based on checkbox checked or not. I have many checkbox but here I am checking address checkbox.
This is my checkbox
<input type="checkbox" id="address">Mark it
here is div I want to hide/show
<div id="options">
//div
</div>
if($('#address').prop('checked')){
$('#options').show()
}else{
$('#options').hide()
}
Above code is not working for me.
2 answers
solution1
3 2021-11-08 05:28:20
Your code will only work on the first time not every time the value of the checkbox changes.
You have to add event listener on change of the checked value of checkbox
$("#address").on("change", e => {
const isChecked = e.target.checked;
if (isChecked) $('#options').show()
else $('#options').hide()
})
if ($('#address').prop('checked')) { $('#options').show() } else { $('#options').hide() } $("#address").on("change", e => { const isChecked = e.target.checked; if (isChecked) $('#options').show() else $('#options').hide() })
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <input type="checkbox" id="address">Mark it <div id="options"> //div </div>
solution2
2 2021-11-08 05:29:11
The reason your code not working, you only check the condition checkbox check or not you have to call the checkbox change event also every time when the checkbox gets clicked. here is a working demo hope it will help you to understand how jquery works.
<!DOCTYPE html> <html> <head> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script> <script> $(document).ready(function () { $('#options').hide(); $('#address').change(function () { if ($('#address').prop('checked')) { $('#options').show() } else { $('#options').hide() } }); }); </script> </head> <body> <div id="options"> //div </div> <input type="checkbox" id="address">Mark it </body> </html>
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