jQuery change event being called twice

Question

I have a form with some input and select boxes, each has class="myClass" . I also have the following script:

$(document).ready(function() {
    $(".myClass").change(function() {
        alert('bla');
    })
});

I dont understand why after each change in select box or input box, this function is being called twice. What's wrong here?

Appreciate your help!

Answer 1

All I can think of is that you used the same class on the form itself. if so, remove the myClass style from your form tag.

Corrected : http://jsfiddle.net/rY6Gq/1/

Faulty one with double alert: http://jsfiddle.net/rY6Gq/

Answer 2

e.stopImmediatePropagation(); is what worked for me.

$(document).ready(function() {
    $(".myClass").change(function(e) {
        e.stopImmediatePropagation();
        alert('bla');
    })
});

Answer 3

Its a bug, You'd add

$("#some_id").unbind('change');

before any change call

Answer 4

It happens when the same class or whatever attribute you are binding also has the same name parent or child. Obviously, when you change a child, parent also gets changed (its child changes). If they have the same class or attribute, it should fire twice. For example, in the following if you bind to "myClass", it will be called twice.

<div class="myclass">
<select class="myClass">   </select>
</div>

Answer 5

if this occurred in IE, it may be this bug as it was for me: http://bugs.jquery.com/ticket/6593

updating to jQuery 1.7.1 worked for me.

Answer 6

change like this

$(document).ready(function() {
    
    $(".myClass").unbind('change');

    $(".myClass").change(function() {
        alert('bla');
    })
});

Answer 7

它不是： http : //jsfiddle.net/nfcAS/

Answer 8

The only one that worked for me was unbind before the change check.

 $(".select2Component").unbind(); 
    $(".select2Component").change(function() { 
         //code
    });

Answer 9

For me - I had written the on change event inside a function. Moving it to $(document).ready(function () {}); solved my case.

Answer 10

Try debugging the code in Firebug rather than alerting. It may be loosing the focus and returning it is causing the appearance of two changes when there isn't two happening