You can't do it by using $.load.
You can do it by using $.ajax, and set the dataType to 'script'
For example:
<div id=box1></div>
<div id=box2></div>
<script type="text/javascript">
$(function () {
$.ajax ({ url:'url.php', dataType: 'script' });
});
PHP Respone:
$('#box1').html('Hello');$('#box2').html('World');
On 23 ינואר, 12:07, Trend-King <i...@trend-king.de> wrote:
> Hello i have a question about getting content via ajax.
>
> i make a ajax call to an seperate php file with will return me the
> content of the to be updated elements on a page. for example <div
> id="box1">the content goes there</div>
> <div id="box2">the second content goes there</div>
>
> how can i manage the ajax result to take these elements on the right
> place in the page.
>
> i want to update each element of the result replace #box1 with the
> #box1 from the ajax call and replace #box2 with the #box2 from the
> ajax call
>
> is this possible or i have to do a call each element?
>
> thanks for your replies
> Jens
