所以我有这个结构
<div class="accordionContent">
<form action="/make_requests" class="make_request" method="post"><div style="margin:0;padding:0;display:inline"><input name="authenticity_token" type="hidden" /></div>
<tr>
<td>Enter Sandman</td>
<td>2</td>
<td class="money">$0.00</td>
<td>
<input class="yes_please" id="group_7,1,2_yes" name="group[7,1,2]" type="radio" value="yes" />
Yes
<input checked="checked" id="group_7,1,2_no" name="group[7,1,2]" type="radio" value="no" />
No
</td>
<tr>
<td>Fade to Black</td>
<td>2</td>
<td class="money">$0.00</td>
<td>
<input class="yes_please" id="group_3,6_yes" name="group[3,6]" type="radio" value="yes" />
Yes
<input checked="checked" id="group_3,6_no" name="group[3,6]" type="radio" value="no" />
No
</td>
</tr>
<tr><td><input id="make_requests" name="commit" type="submit" value="Add to Set" /></td><td><span class="band_notice">Set Added</span></td></tr>
</form>
如果至少有一个单选按钮是yes,我只需要启用输入...任何想法如何在jQuery中执行此操作
答案 0 :(得分:2)
$("input:radio").change(function(){
var checked = $("input:checked");
if ($("input[value='yes']", checked).length > 0)
{ $('#make_requests').attr('disabled',''); }
else
{$('#make_requests').attr('disabled','disabled');}
});
将一个函数绑定到所有复选框的更改事件,获取所有选中的输入,然后计算值为“是”的数量。如果它大于0,则启用提交按钮,如果少则禁用。
答案 1 :(得分:1)
将disabled="disabled"
添加到html中的输入中,并添加这段js:
$('input[type=radio]').change(function() {
$('input[type=submit]').attr('disabled', true);
$('input[type=radio][value=yes]:checked').first().each(function() {
$('input[type=submit]').removeAttr('disabled');
});
});
答案 2 :(得分:1)
这对我有用......
<script type="text/javascript">
$(function () {
$('input:radio').change(function () {
if($('.yes_please:checked').length)
$('#make_requests').removeAttr('disabled');
else
$('#make_requests').attr('disabled', 'disabled');
});
});
</script>