Why does jQuery .change() not fire on radio buttons deselected as a result of a namesake being selected?
Asked Answered
P

3

7

If I have several radio buttons of the same name, only one can be selected at any one time. When one becomes selected, any namesakes lose their selection. I'm intrigued as to why this doesn't constitute a .change() event for jQuery.

Hacking the $.change() function would be pretty drastic, but this doesn't seem to be reported as a bug — so I'm interested in finding out why this isn't the case.

Test below: I'd expect the two change events to fire whenever a radio is selected, but that isn't the case.

Here is the example: http://jsfiddle.net/XzmmW/

Pander answered 18/7, 2011 at 17:26 Comment(0)
A
12

If you think of a set of radio buttons like a select element, it becomes clear why it's unnecessary to fire a change event for the deselected radio button.

When you change the selected option in a select element, you get one change event (you wouldn't expect anything else), rather than one for the previously selected option, and another for the newly selected option.

As has already been mentioned, when one radio button is selected, no others in the same set can be, so it would be redundant to fire a second change event.

Also note that this is not specific to jQuery - the standard JavaScript change event works in exactly the same way.

Alanealanine answered 18/7, 2011 at 17:34 Comment(0)
H
3

There is only 1 change event... you changed from "old selection" to "new selection".

Keep in mind that although you have multiple "elements" they are all part of the same form element (as defined by the name attribute).

Hornbill answered 18/7, 2011 at 17:28 Comment(0)
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2

There is no need to fire an event for each radio button in a named group, when one is selected you know the others are not selected

Blues answered 18/7, 2011 at 17:30 Comment(0)

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