Thanks alot for this.
I'll keep a credit to this page when i use this function.

I´m glad you found it useful.
But I do not require you to credit me for it =)

Not working in FF 1.5.0.6
I calling following
function replaceNodes(){
var myt=document.getElementById('mytable');
var n1 = myt.lastChild.childNodes[0].childNodes[1];
var n2 = myt.lastChild.childNodes[0].childNodes[2];
document.swapNode ? n1.swapNode(n2) : swapNodes(n2,n1);
}

@ Greg
Looks like you are trying to manipulate table-cells, judging from your getElementById('mytable') and Firefox doesn't allow you to do that. It's a limitation in Firefox.

So it's either that or something else that is wrong in your implementation since I also use FF 1.5.0.6 and the site I'm using my function in still works with FF under both Windows and Linux.

Just wanted to say thanks for posting this, really got me out of a hole.

Also worth mentioning that this method (as opposed to the others I've tried) doesn't break any events you might have bound to the elements you are swapping.

Thanks again :-)

@ Richard

I'm glad you found it useful.
I didn't realized that my code doesn't break events but I guess that it makes the function even more useful! =)

@ Greg
This seems to work fine on table ROWS in Firefox, as long as you remember two things:

1. If you don't define one, Firefox always seems to generate a TBODY element as the child of the TABLE and the parent of the rows. I find the easiest workaround is to define a TBODY explicitly and work on the tbody rather than the table

2. If you leave any whitespace between the closing tag and the following , Firefox creates an empty TEXT node. Same problem between the and the first (etc).

Thanks for the code, here's a minor tweak to allow this to work even even if the nodes are not children of the same parent:

function swapNodes(item1,item2) {
// We need a clone of the node we want to swap
var itemtmp = item1.cloneNode(1);

// We also need the parentNodes of the items we are going to swap.
var parent = item1.parentNode;
var parent2 = item2.parentNode;

// First replace the second node with the copy of the first node
// which returns a the new node
item2 = parent2.replaceChild(itemtmp,item2);

//Then we need to replace the first node with the new second node
parent.replaceChild(item2,item1);

// And finally replace the first item with it's copy so that we
// still use the old nodes but in the new order. This is the reason
// we don't need to update our Behaviours since we still have
// the same nodes.
parent2.replaceChild(item1,itemtmp);

// Free up some memory, we don't want unused nodes in our document.
itemtmp = null;
}

a simpliest way is to do
function SwapNode(oParent, ChildNumber)
{
var oRemoved = oParent.removeChild(oParent.childNodes[ChildNumber - 1]);
oParent.insertBefore(oRemoved, oParent.childNodes[ChildNumber]);
}

@ Etienne

That function will only work if the nodes you are trying to swap are next to each other. I suggest using one of the other versions, because this will not always be the case.

Thanks for the function Mattias, much appreciated, saved me a bit of time :)

Also, in IE would it be more efficient (system resource - wise) to use swapNode? or should I stick with this for both IE and FF

Hi Joshua, I'm glad it could help you.
In IE though I would stick to the native swapNode method, it's probably faster also.

Thanks man!!!

Thanks, exactly what I was looking for!.

Amzing solution, Thnx Mattias