开发者

Custom graphical Radio buttons with jQuery is somehow breaking index()

开发者 https://www.devze.com 2023-04-13 04:55 出处:网络
I\'m using this customInput plugin (with jQuery 1.6.2) to customize the look of my radio buttons. It works great.

I'm using this customInput plugin (with jQuery 1.6.2) to customize the look of my radio buttons.

It works great.

The problem I have now is that I'm just trying to get the index() number of the selected radio button and it always returns 0. There are six radio buttons and I'm looking for a number between 0 and 5.

JavaScript:

$('input[name="amount"]').click(function() {
    var x = $(this).filter(':checked').index();
    var y = $(this).filter(':checked').val(); //<-- for troubleshooting
    alert(x + y);  //<-- for troubleshooting
});

Oddly, val() is still working fine and returns the proper 开发者_运维技巧value. Therefore, the form data is always getting the correct radio value.

http://jsfiddle.net/sparky672/LdVGD/

HTML:

<fieldset id="radioset">
    <input type="radio" id="radio-1" name="amount" value="Option 1" checked="checked" /><label for="radio-1" title="">Option 1</label>
    <input type="radio" id="radio-2" name="amount" value="Option 2" /><label for="radio-2" title="">Option 2</label>
    <input type="radio" id="radio-3" name="amount" value="Option 3" /><label for="radio-3" title="">Option 3</label>
    <input type="radio" id="radio-4" name="amount" value="Option 4" /><label for="radio-4" title="">Option 4</label>
    <input type="radio" id="radio-5" name="amount" value="Option 5" /><label for="radio-5" title="">Option 5</label>
    <input type="radio" id="radio-6" name="amount" value="Option 6" /><label for="radio-6" title="">Option 6</label>
</fieldset>

When simply not using the customInput plugin, the index number is then returned.

http://jsfiddle.net/sparky672/LdVGD/1/


Side Question:

After disabling the plugin, the index() returns 0, 2, 4, 6, 8, or 10. It's like the <label> itself is being counted at part of the index(), effectively doubling the count. Why should this be?

I cannot remove the <label> elements as the plugin depends on these to function.

I just want to retrieve a number from 0 to 5 depending on whether radio button 0 through 5 is checked.

Any suggestions? Perhaps another method to check which radio button is selected?

My ultimate goal? To simply change some variables depending on which button is selected.


As you've noticed, your index call doesn't quite work the way you expect it to when you don't use the plugin. From the fine manual:

If no argument is passed to the .index() method, the return value is an integer indicating the position of the first element within the jQuery object relative to its sibling elements.

You're getting values twice as large as you think they should be because the <label> elements are also siblings so you have five <input> elements and five <label> elements in the sibling list.

When you activate the plugin, your radio buttons get wrapped in a <div> so the structure of each button looks like this:

<div>
    <input type="radio">
    <label>
</div>

That leaves the radio button with a single sibling, <label>, and an index of zero.

You already have id attributes on your radio buttons so why not use those to figure out the index? Something like this:

var index = parseInt(this.id.replace(/radio-/, ''), 10) - 1;

Check the third number in these demos:

  • With plugin: http://jsfiddle.net/ambiguous/nGVkC/
  • Without plugin: http://jsfiddle.net/ambiguous/UkPh8/

Or you could use the element form of index:

If .index() is called on a collection of elements and a DOM element or jQuery object is passed in, .index() returns an integer indicating the position of the passed element relative to the original collection.

So you could do it like this:

var i = $('input[name=amount]').index(this);

And some demos:

  • With plugin: http://jsfiddle.net/ambiguous/yAVHn/
  • Without plugin: http://jsfiddle.net/ambiguous/vP3Du/
0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号