That is correct and to add to the solution... I'd recommend caching the
results of the ul selector and leaving out the 'div' part unless it is
necessary.
var $img = $('.gallery img'), $ul = $('#bild_spel ul');
$img.each(function() {
    $ul.append(this);
});

--
Brandon Aaron


On Mon, Mar 30, 2009 at 6:52 AM, Gilles <gil...@netxtra.net> wrote:

>
> For some reason your answer doesn't show on here. You might have mail
> me directly, anyway if you want to insert the element straight away
> instead of creating a new image element maybe the fact that you are
> mixing string variable and an object in your appen() might cause the
> problem.
>
> In your code if you alert this you will see that it isn't <img
> src="......"> but an object, an image tag object, so i think your code
> fail because you are tying to append a string and an object together.
>
> this is not a string. Try appending this on its own and see which
> result it gives.
>
> This should work (also wont be wrapped in <li>)
>
> var img = $(".gallery img");
> $(img).each(function() {
>         $("#bild_spel div ul").append(this);
>
> });
>
> You'll just need to wrap it in your <li> somewhere else
>
> On Mar 30, 12:17 pm, Gilles <gil...@netxtra.net> wrote:
> > Might want to try something like this instead:
> >
> > $(".gallery img").each(function()
> > {
> >       $('#bild_spel div ul').append('<li><img src=" '+$(this).attr
> > ('src')+' " alt=""></li>');
> >
> > });
> >
> > haven't tested it but I believe it's correct :)
> >
> > On Mar 30, 11:48 am, smurkas <marcus.dalg...@gmail.com> wrote:
> >
> > > Hello.
> >
> > > I have a situation where I have grabbed a couple of images off a page
> > > and want to insert them again into an unordered list. Grabbing the
> > > images was no problem but I don't know what to write when inserting
> > > them again.
> > > The code below does not result in the images being inserted again,
> > > instead each list item contains
> > > [object HTMLImageElement] as text. I also tried wrapping this in $
> > > (this) but that didn't work either. The solution is probably dead
> > > simple but right now I am missing it. Can anyone help me out?
> >
> > > var img = $(".gallery img");
> > > $(img).each(function() {
> > >         $("#bild_spel div ul").append('<li>'+this+'</li>');
> >
> > > });
> >
> > > Kindly, Marcus.
> >
>

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