如何使用 JQuery 侦听 RadioGroup 选定值的更改？
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How do I listen for changes to a RadioGroup's selected value using JQuery?
提问by Justin
I need to register a handler for a group of radio buttons. I'm using JQuery and hoped that its .changemethod would accomplish this. However I have not experienced the desired behavtheitroad.
我需要为一组单选按钮注册一个处理程序。我正在使用 JQuery 并希望它的.change方法能够实现这一点。但是，我还没有经历过想要的行为。
Here is a sample snippet I've written. Sadly, the "radioValueChanged" is only called on the initial load. Selecting either true / false does not trigger the handler.
这是我写的一个示例片段。遗憾的是，“radioValueChanged”仅在初始加载时被调用。选择 true / false 不会触发处理程序。
<html>
<script src="jquery-1.4.2.min.js" type="text/javascript"></script>
<form id="myForm">
<div id="Question1Wrapper">
<div>
<input type="radio" name="controlQuestion" id="valueFalse" value="0" />
<label for="valueFalse">
False</label>
</div>
<div>
<input type="radio" name="controlQuestion" id="valueTrue" value="1" />
<label for="valueTrue">
True</label>
</div>
</div>
<div id="Question2Wrapper">
<div>
<label for="optionalTextBox">
This is only visible when the above is true</label>
<input type="text" name="optionalTextBox" id="optionalTextBox" value="" />
</div>
</div>
<script type="text/javascript">
jQuery(document).ready(function ()
{
$("#controlQuestion").change(radioValueChanged('controlQuestion'));
})
function radioValueChanged(radioName)
{
radioValue = $('input[name=' + radioName + ']:checked', '#myForm').val();
alert(radioValue);
if(radioValue == 'undefined' || radioValue == "0")
{
$('#Question2Wrapper:visible').hide();
}
else
{
$('#Question2Wrapper:visible').show();
}
}
</script>
</form>
回答by Quintin Robinson
There are a few issues here.
这里有几个问题。
You are immediately running
radioValueChanged('controlQuestion')upon script execution because that is a method call and not a function assignment.The selector
$("#controlQuestion")is wrong, you don't have any elements with id ofcontrolQuestion.The
radioValueChangedmethod is not properly handling values as they would be passed to a jQuery event handler.
您会
radioValueChanged('controlQuestion')在脚本执行时立即运行，因为那是方法调用而不是函数分配。选择器
$("#controlQuestion")是错误的，您没有任何 id 为 的元素controlQuestion。该
radioValueChanged方法未正确处理值，因为它们将传递给 jQuery 事件处理程序。
You could try something like the following:
您可以尝试以下操作：
jQuery(document).ready(function ()
{
$("input[name='controlQuestion']").change(radioValueChanged);
})
function radioValueChanged()
{
radioValue = $(this).val();
alert(radioValue);
if($(this).is(":checked") && radioValue == "0")
{
$('#Question2Wrapper').hide();
}
else
{
$('#Question2Wrapper').show();
}
}
In all honesty I'm not sure if that is the actual logic you are looking for with the if statement, but hopefully this will provide a basis for you to correct the current code.
老实说，我不确定这是否是您使用 if 语句寻找的实际逻辑，但希望这将为您更正当前代码提供基础。
回答by John Strickler
$('#Question2Wrapper:visible').show();
Take out the :visible, this will only select it if the div is already showing, in effect it will never be shown if it is hidden.
取出:visible，这只会在 div 已经显示时选择它，实际上，如果它被隐藏，它将永远不会显示。
$('#Question2Wrapper').show();
I digress, I think Quintin hits most of the points. There are a few issues going on here.
我离题了，我认为昆汀击中了大部分分数。这里有几个问题。
