Question

The following line This does NOT work in my code in firefox/firebug (but works fine in JSFIDDLE), you will see below I have a work around, just wondering if anyone knows the internal reason why?

var checkedVal = parseInt($('input[@name=' + uniqueNamePart + 'currDim]:checked').val(), 10);

http://jsfiddle.net/darrenshrwd/eqh2y/43/

<input type="radio" value="1" name="blahcurrDim">One
<input type="radio" checked="" value="2" name="blahcurrDim">Two
<input type="radio" value="3" name="blahcurrDim">Three
<input type="radio" value="4" name="blahcurrDim">Four

​…

$('document').ready(

function() {

    var uniqueNamePart = "blah";

    var dimensionClick = function() {

        // This does NOT work in my code in firefox/firebug (but works fine in JSFIDDLE):
        var checkedVal = parseInt($('input[@name=' + uniqueNamePart + 'currDim]:checked').val(), 10);

        // This does work in both:
        //var myRadio = $('input[name=' + uniqueNamePart  + 'currDim]'),
        //    checkedVal = parseInt(myRadio.filter(':checked').val(), 10);        

        alert(checkedVal);

    };

    $('input[name=' + uniqueNamePart + 'currDim]:radio').click(dimensionClick);

});

Answer 1

Your attribute selector is malformed. An @ character is not necessary:

var checkedVal
    = parseInt($('input[name=' + uniqueNamePart + 'currDim]:checked').val(), 10);

That code appears to work in your fiddle, but that’s because the only <input> elements there are the radio buttons, so matching succeeds even if the invalid attribute selector is ignored.