0

我正在尝试移动带有stamp2 类的元素，所以它在第五个孩子之后。这是我尝试的。该项目似乎已删除。

$('.stamp2').remove().after('.section:nth-child(5)');

html:

<section class="photo small stamp stamp1">..</section>
<section class="photo small stamp stamp2">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
<section class="photo small">..</section>
4

4 回答 4

5

删除该.remove()位，只需使用after()它将移动节点：

演示

$('section:nth-child(5)').after($('.stamp2'))
于 2013-08-28T11:59:58.083 回答
1

尝试.insertAfter()

$('.stamp2').detach().insertAfter('.section:nth-child(5)')
于 2013-08-28T12:00:32.147 回答
1

你不需要删除它。只需使用 after 功能，因为它会做的伎俩

$('.stamp2').after($('.section:nth-child(5)'));
于 2013-08-28T12:03:15.327 回答
0

使用 Detach() 而不是 Remove() 然后使用 InsertAfter()

于 2013-08-28T12:08:59.600 回答