An alternative would be to call cloneNode() for each DOM node before
appending to the DOM tree. Here is a small untested function to
illustrate:
function appendChildNodeClones(parent, /* ... */) {
var args = flattenArguments(arguments);
for (var i = 1; i < args.length; i++) {
if (args[i] != null && typeof(args[i].cloneNode) == "function") {
parent.appendChild(args[i].cloneNode(true));
} else {
appendChildNodes(parent, args[i]);
}
}
}
/Per
On Mon, Apr 28, 2008 at 1:40 AM, Christoph Zwerschke <[EMAIL PROTECTED]> wrote:
>
> machineghost schrieb:
>
> > It turns out the problem is with both of our understanding of repeat
>
> Ah, yes! Thanks for your analysis, now everything makes sense.
>
> I could use the excuse that I usually program in Python, not in
> Javascript, but in fact repeat() does not behave differently there ;-)
>
> To summarize, the Test2 code based on two subtle misconceptions. First,
> the implicit idea that in the expression repeat(BR()) the BR() is either
> evaluated on every repetition, or copies of the BR() object are created
> on every repetition, which both is of course not the case. Second, the
> idea that you don't need to care whether elements you're adding to the
> DOM are already used elsewhere ;-)
>
> Your solution
>
>
> var brAdder = function(x) { return [x, BR()]; }
> appendChildNodes("test2", map(brAdder, text));
>
> is nice. Alternatively, the Test2 code be fixed as follows:
>
> appendChildNodes("test2", izip(text, imap(BR, repeat())));
>
> Just to make use of repeat() against these odds...
>
>
>
> -- Christoph
>
> >
>
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