new_Url:
<td class="posts column-posts"><a href="edit.php?tshowcase-categories=ops&post_type=tshowcase">2</a></td>old_url:
<span class="view"><a href="https://blog.company.com/team/tshowcase-categories/ops/">View</a></span>我想用new_url替换旧的url。
var new_url = $("td.column-posts").find("a").attr('href');
$("span.view").find("a").attr('href',new_url)我无法获取new_url的值。
PLease帮助。谢谢
下面是New_url的完整结构:
<tbody id="the-list" data-wp-lists="list:tag">
<tr id="tag-7" class="alternate">
<td class="posts column-posts"><a href="edit.php?tshowcase-categories=ops&post_type=tshowcase">1</a></td>
</tr>
</tbody>发布于 2014-02-27 13:49:56
您可以这样做,只需创建链接的ids:
$('#2').attr('href', $('#1').attr('href'));这是小提琴。http://jsfiddle.net/Goodluck/PasNk/1/
发布于 2014-02-27 14:26:31
将表体包含在表标记内
<table>
<tbody id="the-list" data-wp-lists="list:tag">
<tr id="tag-7" class="alternate">
<td class="posts column-posts"><a href="edit.php?tshowcase-categories=ops&post_type=tshowcase">1</a>
</td>
</tr>
</tbody>
</table>Working Fiddle
发布于 2014-02-27 13:51:47
如果只有一个具有column-posts类的td
var new_url = $("td.column-posts a").attr('href');
$("span.view a").attr('href',new_url)如果有多个.column-posts td,则必须使用each编写,如下所示
$("td.column-posts a").each(function(){
$("span.view a").attr('href',new_url) //you have to map each span to each a
})https://stackoverflow.com/questions/22060004
复制相似问题