乘法无线电和选择功能在safari中不起作用(Multiplication radio and select function doesn't work in safari)

乘法无线电和选择功能在safari中不起作用(Multiplication radio and select function doesn't work in safari)

出于某种原因，这段代码在Safari中不起作用，任何人都有任何线索为什么？ 第一个收音机盒很好，但是一旦我检查了第二个收音机盒，它就不会计算出来。 http://jsfiddle.net/njyEP/

谢谢

使用Javascript：

$("input[type=radio][name=radio],select").change(function() { var radio = parseFloat($('#radio:checked').attr("tvalue")); var select = parseFloat($('#select').val()); $('#Total').text((select * radio)); });

HTML

<form method="post" action=""> <select name="select" id="select" name="select"><option value="1">1</option><option value="2">2</option><option value="3">3</option><option value="4">4</option><option value="5">5</option><option value="6">6</option><option value="7">7</option><option value="8">8</option><option value="9">9</option><option value="10">10</option> </select> <div id="radioButtons"><input type="radio" class="multi" name="radio" id="radio" ivalue="1" tvalue="275" value="1185">Earlybird 1 Day Registration $275<br> <input type="radio" class="multi" name="radio" id="radio" ivalue="2" tvalue="475" value="1186">Earlybird 2 Day Registration $475<br> <input type="radio" class="multi" name="radio" id="radio" ivalue="3" tvalue="625" value="1187">Earlybird 3 Day Registration (all three days) $625<br> </div><div id="checkBoxes"> <br><label id="Total"></label><br></form>

For some reason this code doesn't work in Safari anyone have any clue to why? The first radio box is fine but as soon as I check the second radio box it doesn't calculate. http://jsfiddle.net/njyEP/

Thanks

Javascript:

$("input[type=radio][name=radio],select").change(function() { var radio = parseFloat($('#radio:checked').attr("tvalue")); var select = parseFloat($('#select').val()); $('#Total').text((select * radio)); });

html

<form method="post" action=""> <select name="select" id="select" name="select"><option value="1">1</option><option value="2">2</option><option value="3">3</option><option value="4">4</option><option value="5">5</option><option value="6">6</option><option value="7">7</option><option value="8">8</option><option value="9">9</option><option value="10">10</option> </select> <div id="radioButtons"><input type="radio" class="multi" name="radio" id="radio" ivalue="1" tvalue="275" value="1185">Earlybird 1 Day Registration $275<br> <input type="radio" class="multi" name="radio" id="radio" ivalue="2" tvalue="475" value="1186">Earlybird 2 Day Registration $475<br> <input type="radio" class="multi" name="radio" id="radio" ivalue="3" tvalue="625" value="1187">Earlybird 3 Day Registration (all three days) $625<br> </div><div id="checkBoxes"> <br><label id="Total"></label><br></form>

最满意答案

您不应该定义具有相同ID的多个元素。 要使您的示例在Safari中工作，请将您的第二个javascript行更改为

var radio = parseFloat($('input[type=radio]:checked').attr("tvalue"));

http://jsfiddle.net/2TCGp/

（或者，如果你遵循devnull69的建议并使用data-tvalue，你可以使用jQuery的.data（）方法）。

You shouldn't define multiple elements with the same id. To make your example work in Safari, change your second javascript line to

var radio = parseFloat($('input[type=radio]:checked').attr("tvalue"));

http://jsfiddle.net/2TCGp/

(Or, if you follow devnull69's advice and use data-tvalue, you can use jQuery's .data() method).