假设我们有以下代码。单击链接时,应发送 ajax 请求。接收数据可能会花费很多时间,期间很容易有时间再点击 1-2 个链接。鉴于 $.ajax 异步工作,问题出现了。但是,如何通过单击加载数据的方式将指向元素的链接传递给错误和成功回调?
$(function(){
$('table tr').on('click', 'td>a', function(){
$(this).parent().parent().after('<tr>'
+'<td>Additional row</td>'
+'<td>Some data...</td>'
+'<td>Some data...</td>'
+'</tr>');
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<td><a href="#">Load additional rows</a></td>
<td>Some data...</td>
<td>Some data...</td>
</tr>
<tr>
<td><a href="#">Load additional rows</a></td>
<td>Some data...</td>
<td>Some data...</td>
</tr>
<tr>
<td><a href="#">Load additional rows</a></td>
<td>Some data...</td>
<td>Some data...</td>
</tr>
<tr>
<td><a href="#">Load additional rows</a></td>
<td>Some data...</td>
<td>Some data...</td>
</tr>
</tbody>
</table>
1 个回答