views:

72

answers:

1

I have a page with several sets of radio buttons that are used to set options. When one clicks on specific ones, others are selected by default using click event handlers. The functionality works perfectly, but there is an issue with the button's visual state.

I'm using jQueryUI's .buttonset() method to improve the aesthetics, and when I trigger a .click() event programatically, the button does not change state visually. This can result in the current options being quite different from what appears on screen.

Sample code to illustrate the problem:

<fieldset>
    <label for="button1">Button 1</label>
    <input type="radio" id="button1" name="test" />

    <label for="button2">Button 2</label>
    <input type="radio" id="button2" name="test" />
</fieldset>

$('fieldset').buttonset();

$('#button2').click(function() {
    alert('button 2 clicked');
});

$('#button2').click();

​I also set up a fiddle so you can see it in action, if you so desire: http://jsfiddle.net/T5MGh/

As you would expect, the alert box pops up on page load as it should, but the button does not change visually as it does from a user-click.

Any thoughts?

+1  A: 

You can click the actual label that the button set uses, like this:

$('[for=button2]').click();

This works because your structure looks like this after .buttonset():

<fieldset class="ui-buttonset">
    <label for="button1" aria-pressed="false" class="ui-button ui-widget ui-state-default ui-button-text-only ui-corner-left" role="button" aria-disabled="false"><span class="ui-button-text">Button 1</span></label>
    <input type="radio" id="button1" name="test" class="ui-helper-hidden-accessible">

    <label for="button2" aria-pressed="true" class="ui-button ui-widget ui-state-default ui-button-text-only ui-corner-right ui-state-active ui-state-hover" role="button" aria-disabled="false"><span class="ui-button-text">Button 2</span></label>
    <input type="radio" id="button2" name="test" class="ui-helper-hidden-accessible">
</fieldset>

It doesn't work initially because of how jQuery UI does it, it relies on the click coming through the <label>, and the defult browser behavior actually clicking the <input> from that.

Nick Craver
Works like a charm. Thanks! I would not have figured that out quickly without help.
Rookwood
@Rookwood - Welcome :)
Nick Craver
As a note for anyone who reads this in the future, it seems that it is still necessary to .click() on the radio button, itself, as well.$('[for=button_id], #button_id').click();Otherwise, it was changing visually, but not affecting the actual button state, the opposite of the original problem.
Rookwood