下面一行这在我的firefox/firebug代码中不起作用(但在JSFIDDLE中效果很好),你会看到下面我有一个解决方案,只是想知道是否有人知道内部原因?
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);
};
$('input[name=' + uniqueNamePart + 'currDim]:radio').click(dimensionClick);
});
您的属性选择器格式不正确。不需要@
字符:
var checkedVal
= parseInt($('input[name=' + uniqueNamePart + 'currDim]:checked').val(), 10);
该代码似乎在你的小提琴中工作,但这是因为那里唯一的<input>
元素是单选按钮,所以即使忽略无效的属性选择器,匹配也会成功。