如何使用Ajax响应取代HTML元素? [英] How to replace html element with ajax response?
问题描述
如何更换从Ajax响应的HTML元素?我所知道的就是要删除的元素,我怎么替换删除标签与Ajax响应? 例如:
我有code是这样的:
< UL ID =产品>
...............
< / UL>
当我点击一个按钮,AJAX调用到codeginter控制器在那里我收到来自数据库的新数据拉升,而在另一个视图,它开始从UL和结束于关闭UL呈现。
在阿贾克斯成功的功能我这样做:
$('#产品)删除();
//做什么,现在这里与阿贾克斯的响应更换去除部分?
您可以使用replaceWith(参见: HTTP://api.jquery .COM / replaceWith / )
我爱: $('#产品)replaceWith(响应);
How do I replace the html element from ajax response? What I know to is to remove the element, how do I replace that removed tag with ajax response? For example:
I have code like this:
<ul id="products">
...............
</ul>
When I click on a button the ajax call is made to codeginter controller where I recieve the new data pulled from the database and rendered in another view which starts from ul and ends at closing ul.
In ajax success function I do this:
$('#products').remove();
//What to do now here to replace the removed portion with response of ajax?
You can use replaceWith (see: http://api.jquery.com/replaceWith/)
Like: $('#products').replaceWith(response);
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