Question

var random1 = Math.floor((Math.random()*elems.length));
var random2 = Math.floor((Math.random()*elems.length));

while (random1 == random2) {
    var random2 = Math.floor((Math.random()*elems.length));
}

$j('#project-scroller').append( elems );
$j('#project-scroller').append( elems[random1] );
$j('#project-scroller').append( elems[random2] );

So here’s what’s happening: elems gets appended but elms[random1] and elems[random2] doesn’t.

So then I tried this instead:

$j('#project-scroller').append( elems );
$j('#project-scroller').append( elems );

And elems was only appended once, that time, too! So then I tried this instead:

$j('#project-scroller').append( elems[random1] );
$j('#project-scroller').append( elems[random2] );

And both random elems were appended. Then I tried appending random1 twice, and it was only appended once…

What am I missing? Why won’t jQuery append the same elements more than once? Tried the same thing with appendTo (just for kicks), and it does the same thing.

---

Thanks for the answer! Here’s what I ended up with:

var random1 = Math.floor((Math.random()*elems.length));
var random2 = Math.floor((Math.random()*elems.length));
while (random1 == random2) {
    var random2 = Math.floor((Math.random()*elems.length));
}
newElem1 = $j(elems[random1]).clone();
newElem2 = $j(elems[random2]).clone();
$j('#project-scroller').append( elems ).append( newElem1 ).append( newElem2 );

Answer 1

It’s not appending a copy of elems, it’s appending the real thing. Use clone() to create a 2nd copy to append.