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下面这行不能在firefox/firebug中的代码中工作(但在JSFIDDLE中工作正常),您将在下面看到我有一个解决方法,只是想知道是否有人知道内部原因是什么?为什么这个jQuery输入选择器需要过滤器解决方法
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);
});
完全正确...这就是我从复制另一个stackoverflow问题所得到的结果。 – Darren 2012-07-26 06:54:59