Another thing you can do instead of giving every link a class, you can
give the container a class and select the links in that container.
example html:
<div class="menu">
<ul>
<li><a href="test1">test1</a></li>
<li><a href="test2">test2</a></li>
<li><a href="test3">test3</a></li>
<li><a href="test4">test4</a></li>
</li>
</div>
use this selector: jQuery('.menu a')
That way you will get every link inside the div with the class "menu"
You probably already have a class indicating the menu position for the
css.. you can just use that selector
On Jun 26, 11:40 pm, Pegpro <[EMAIL PROTECTED]> wrote:
> I am currently using:
> jQuery.noConflict();
> jQuery(document).ready(function(){
> jQuery(\"a:contains('Home')\").click(function(e){
> e.preventDefault();
> var link = jQuery(this), link =
> link.attr('href');
> jQuery('#loading').fadeIn(200)
> jQuery('.wrap').slideUp(600, function(){
> setTimeout('window.location =
> link', 500);
> });
> });
> Instead of duplicating this for everylink in the menu is there a way I
> can use a or statement within this line:
> jQuery(\"a:contains('Home')\").click(function(e){
>
> Thanks for your help in advance!
