jQuery $(“#radioButton”).change(…) not firing during de-selection

About a month ago Mitt’s question went unanswered. Sadly, I’m running into the same situation now.


Here’s the situation: I’m using jQuery to capture the changes in a radio button. When the radio button is selected I enable an edit box. When the radio button is de-selected, I would like the edit box to be disabled.

The enabling works. When I choose a different radio button in the group, the change event is not fired. Does anyone know how to fix this?

<input type="radio" id="r1" name="someRadioGroup"/> 

<script type="text/javascript">
    $("#r1").change(function () {
        if ($("#r1").attr("checked")) {
        else {
            $('#r1edit:input').attr('disabled', true);



Looks like the change() function is only called when you check a radio button, not when you uncheck it. The solution I used is to bind the change event to every radio button:

$("#r1, #r2, #r3").change(function () {

Or you could give all the radio buttons the same name:

$("input[name=someRadioGroup]:radio").change(function () {

Here's a working jsfiddle example (updated from Chris Porter's comment.)

Per @Ray's comment, you should avoid using names with . in them. Those names work in jQuery 1.7.2 but not in other versions (jsfiddle example.).


You can bind to all of the radio buttons at once by name:


Working example here: http://jsfiddle.net/Ey4fa/


This normally works for me:

if ($("#r1").is(":checked")) {}


My problem was similar and this worked for me:

$('body').on('change', '.radioClassNameHere', function() { ...

Let's say those radio buttons are inside a div that has the id radioButtons and that the radio buttons have the same name (for example commonName) then:

$('#radioButtons').on('change', 'input[name=commonName]:radio', function (e) {
    console.log('You have changed the selected radio button!');

Same problem here, this worked just fine:

$('input[name="someRadioGroup"]').change(function() {
    $('#r1edit:input').prop('disabled', !$("#r1").is(':checked'));

With Ajax, for me worked:


<div id='anID'>
 <form name="nameOfForm">
            <p><b>Your headline</b></p>
            <input type='radio' name='nameOfRadio' value='seed' 
             <?php if ($interviewStage == 'seed') {echo" checked ";}?> 
            onchange='funcInterviewStage()'><label>Your label</label><br>


 function funcInterviewStage() {

                var dis = document.nameOfForm.nameOfRadio.value;

                //Auswahltafel anzeigen
                  if (dis == "") {
                      document.getElementById("anID").innerHTML = "";
                  } else { 
                      if (window.XMLHttpRequest) {
                          // code for IE7+, Firefox, Chrome, Opera, Safari
                          xmlhttp = new XMLHttpRequest();
                      } else {
                          // code for IE6, IE5
                          xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
                      xmlhttp.onreadystatechange = function() {
                          if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                              document.getElementById("anID").innerHTML = xmlhttp.responseText;

And php:

//// Get Value
$id = mysqli_real_escape_string($db, $_GET['id']);

//// Insert to database
$insert = mysqli_query($db, "UPDATE [TABLE] SET [column] = '$id' WHERE [...]");

//// Show radio buttons again
$mysqliAbfrage = mysqli_query($db, "SELECT [column] FROM [Table] WHERE [...]");
                  while ($row = mysqli_fetch_object($mysqliAbfrage)) {
                  <form name='nameOfForm'>
                      <p><b>Your headline</b></p>
                      <input type='radio' name='nameOfRadio' value='seed'"; if ($interviewStage == 'seed') {echo" checked ";} echo" onchange='funcInterviewStage()'><label>Yourr Label</label><br>
                      <input type='radio' name='nameOfRadio' value='startup'"; if ($interviewStage == 'startup') {echo" checked ";} echo" onchange='funcInterviewStage()'><label>Your label</label><br>

                  </form> ";
<input id='r1' type='radio' class='rg' name="asdf"/>
<input id='r2' type='radio' class='rg' name="asdf"/>
<input id='r3' type='radio' class='rg' name="asdf"/>
<input id='r4' type='radio' class='rg' name="asdf"/><br/>
<input type='text' id='r1edit'/>                  

jquery part

$(".rg").change(function () {

if ($("#r1").attr("checked")) {
        else {
            $('#r1edit:input').attr('disabled', 'disabled');

here is the DEMO


The change event not firing on deselection is the desired behaviour. You should run a selector over the entire radio group rather than just the single radio button. And your radio group should have the same name (with different values)

Consider the following code:

$('input[name="job[video_need]"]').on('change', function () {
    var value;
    if ($(this).val() == 'none') {
        value = 'hide';
    } else {
        value = 'show';

I have same use case as yours i.e. to show an input box when a particular radio button is selected. If the event was fired on de-selection as well, I would get 2 events each time.


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