Richard,
Thanks for your reply.
It is neat but it leaks memory on IE.
While the following not:
$.fn.replaceWithAndReturnNew = function(html){
var div = document.createElement('div');
var replaced = this[0];
replaced.parentNode.replaceChild(div, replaced);
div.innerHTML = html;
var replacer = div.firstChild;
div.parentNode.replaceChild(replacer, div);
return $(replacer);};
The trick here is to insert the HTML after placing the element in the
DOM.
I've set a 1000 time loop over the example 4 at the page
http://beebole.com/pure/getting_started.html#example4
And with IE it's clear the memory is climbing fast.
Not sure this can be a bug for jQuery, as we are a bit extreme here on
the innerHTML usage.
Cheers,
On Sep 20, 2:31 pm, "Richard D. Worth" <[EMAIL PROTECTED]> wrote:
> $.fn.replaceWithAndReturnNew = function(htmls){
> var replacer = $(htmls);
> $(this).replaceWith(replacer);
> return replacer;
>
> };
>
> - Richard
>
> On Sat, Sep 20, 2008 at 3:31 AM, tchvil <[EMAIL PROTECTED]> wrote:
>
> > Thanks a lot for your replies.
> > Ok, I guess there are good reasons to return the replaced object.
>
> > Now my question becomes... is there a better jQuery way of doing this
> > quite ugly thing:
>
> > $.fn.replaceWithAndReturnNew = function(htmls){
> > var replaced = $(this)[0];
> > var div = document.createElement('DIV');
> > div.innerHTML = htmls;
> > var replacer = div.firstChild;
> > replaced.parentNode.replaceChild(replacer, replaced);
> > return $(replacer);};
>
> > There's a working page there:http://friendpaste.com/GkTTkwQV
>
> > Cheers,
>
> > On Sep 18, 1:09 am, Ariel Flesler <[EMAIL PROTECTED]> wrote:
> > > Sorry, I missunderstood the initial (and subsequential) posts.
> > > The present behavior (return the initially matched element) is the
> > > correct one. That's what I'd expect and it matches the behavior of
> > > remove(), append(), etc.
>
> > > Don't file any bug report :)
>
> > > Thanks Richard for pinging me about this.
>
> > > --
> > > Ariel Fleslerhttp://flesler.blogspot.com/
>
> > > On Sep 17, 6:12 pm, Ariel Flesler <[EMAIL PROTECTED]> wrote:
>
> > > > I think this is a bug, I'd expectreplaceWithto return the old set,
> > > > just like append/etc.
>
> > > > Can you (tchvil) file a bug with a test case ? put it under core,
> > > > assigned to me.
>
> > > > Thanks
> > > > --
> > > > Ariel Fleslerhttp://flesler.blogspot.com/
>
> > > > On Sep 17, 8:04 am, "Richard D. Worth" <[EMAIL PROTECTED]> wrote:
>
> > > > > On Wed, Sep 17, 2008 at 6:58 AM, Karl Rudd <[EMAIL PROTECTED]>
> > wrote:
>
> > > > > > Nah, it's not a bug.
>
> > > > > >replaceWith( content ) Returns: jQuery
> > > > > > Replaces all matched elements with the specified HTML or DOM
> > elements.
> > > > > > This returns the JQuery element that was just replaced, which has
> > been
> > > > > > removed from the DOM
>
> > > > > > (http://docs.jquery.com/Manipulation)
>
> > > > > It may be well documented, but that doesn't mean it can't be a bug.
> > ;)
>
> > > > > > It's like doing a remove(), followed by an append(). Like
> > "remove()"
> > > > > > it also "returns" (well keeps in the jQuery collection) the
> > elements
> > > > > > it removes.
>
> > > > > This is quite a fundamental difference. Since .remove() returns a DOM
> > > > > Element, it's not chainable. SincereplaceWith() returns a chainable
> > jQuery
> > > > > object, it's reasonable to expect methods you call on that chain to
> > effect
> > > > > the new element, not your now removed/disconnected element(s). Like
> > many
> > > > > other methods, it could modify the chain. And if you wanted to get
> > back to
> > > > > the previous chain, you could call .end(). My 2c.
>
> > > > > - Richard
