radio button to enable form fields
I have a form containing two radio buttons, I want to enable corresponding fields when a radio button is checked. My code just isn't working, here it is:
<head>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function() {
var $type = $('select'),
$field1 = $("#fieldID1"),
$field2 = $("#fieldID2");
$type.change(function() {
var isDisabled = ($type.filter(":checked").val() === "1");
$field1.prop("disabled", !isDisabled);
开发者_运维知识库 $field2.prop("disabled", isDisabled);
}).change();
});
</script>
</head>
My html body is like:
<body>
<form action="" method="post">
<input type="radio" name="select" value="1" />
<input type="text" name="fieldID1" id="fieldID1"/>
<input type="radio" name="select" value="2" />
<table id="fieldID2">
<tr>...</tr>
<tr>...</tr>
...
</table>
<input type="submit" value="Add">
</form>
</body>
I referenced the code from this post. Can anyone point me out what the problem is, I am not quite familiar with jQuery, thanks!
Still cannot be fixed...
$fieldID1.prop("disabled", !isDisabled);
$fieldID2.prop("disabled", isDisabled);
Should probably be:
$field1.prop("disabled", !isDisabled);
$field2.prop("disabled", isDisabled);
$(document).ready(function() {
var $type = $(':radio[name="select"]'),
$field1 = $("#fieldID1"),
$field2 = $("#fieldID2");
$type.change(function() {
var isDisabled = ($type.filter(":checked").val() === "1");
$field1.prop("disabled", !isDisabled);
$field2.prop("disabled", isDisabled);
}).change();
});
加载中,请稍侯......
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