这是代码
<table class="table">
<tbody>
<tr>
<td></td>
<td>Destination</td>
<td>Service Charges</td>
<td>Time Frame</td>
</tr>
<tr class="uk">
<td><input checked="checked" type="radio" name="shiping-method" id="shipping-method" value="1"></td>
<td>uk</td>
<td>free</td>
<td>2-4 working days</td>
</tr>
<tr class="uk">
<td><input type="radio" name="shiping-method" id="shipping-method" value="3"></td>
<td>uk</td>
<td>6.00</td>
<td>next working day</td>
</tr>
</tbody>
</table>
主页上无法访问这些值 我正在尝试使用以下代码
$('input[type="radio"]').on('click',function(){
var value = $(this).val();
alert($(this).val());
});
答案 0 :(得分:0)
您可以通过以下方式使用event delegation: -
$('.table').on('click','input[type="radio"]',function(){
var value = $(this).val();
alert($(this).val());
});
工作示例: -
$('.table').on('click','input[type="radio"]',function(){
var value = $(this).val();
alert($(this).val());
});
$('.clickMe').click(function(){
$("<tr class='uk'><td><input checked='checked' type='radio' name='shiping-method' id='shipping-method' value='1'></td><td>uk</td><td>free</td><td>2-4 working days</td></tr><tr class='uk'><td><input type='radio' name='shiping-method' id='shipping-method' value='3'></td><td>uk</td><td>6.00</td><td>next working day</td></tr>").insertAfter($('.table tbody tr'));
});
table {
width: 100%;
}
tr {
height: 50px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table class="table">
<tbody>
<tr>
<td>Check</td>
<td>Destination</td>
<td>Service Charges</td>
<td>Time Frame</td>
</tr>
</tbody>
</table>
<br>
<input type="button" class="clickMe" value="ClickMe!">