Thanks for all the help folks - problem sorted out now - I appreciate
it. Take care
Alex
On 9 June, 16:11, mkmanning <michaell...@gmail.com> wrote:
> For the example markup you give it's very simple, just do this:
>
> $('ul li a').each(function(){
> $(this).replaceWith( $(this).children() );
>
> });
>
> On Jun 9, 6:26 am, Adardesign <adardes...@gmail.com> wrote:
>
> > Run a .each() to retrieve the images, and delete .remove()
> > the images and place them into a var.
>
> > Then remove all li gt(0) greater then and replace them into one <li>
>
> > It should be doable...
>
> > On Jun 9, 9:07 am, Armand Datema <nok...@gmail.com> wrote:
>
> > > mm i dont know if this is the best way but you could try
>
> > > $("ul a#click").onclick = function() { return false; }
>
> > > and then on that link you set a style class with the standard cursor
>
> > > Its a workaround but it works
>
> > > On Tue, Jun 9, 2009 at 2:13 PM, alex <alex_bille...@hotmail.com> wrote:
>
> > > > Hi folks
>
> > > > This has me stumped. I have a list of images and I want to remove the
> > > > links but retain the images.
>
> > > > So this
>
> > > > <ul>
> > > > <li><a href="foo"><img src="path/to/image1" /></a></li>
> > > > <li><a href="foo"><img src="path/to/image2" /></a></li>
> > > > etc etc
> > > > </ul>
>
> > > > Becomes this
>
> > > > <ul>
> > > > <li><img src="path/to/image1" /></li>
> > > > <li><img src="path/to/image2" /></li>
> > > > </ul>
>
> > > > Using this
>
> > > > $('ul a').remove();
>
> > > > does not work as it also removes the images as well.
>
> > > > Any help would be much appreciated. Thanks in advance.
>
> > > > Alex
>
> > > --
> > > Armand Datema
> > > CTO SchwingSoft
