Wow, 39 minutes for self-filing and self-fixing ?! Is this a
record ;o)
--DBJ
On Jul 15, 12:50 am, John Resig <jere...@gmail.com> wrote:
> This one was tricky because we've never explicitly said *not* to use
> multiple elements in wrap, just that you should have one. There were two
> options:
> - Ignore the remaining elements (as you suggested) and possibly break some
> unknown code.
> - Create a new case where .pushStack is used, potentially breaking code
> that depends on normal cases like .wrap("<p></p>") behaving.
>
> It seemed like #2 was much more likely to cause problems so I opted for your
> patch. Filed and
> fixed:http://dev.jquery.com/ticket/4903http://dev.jquery.com/changeset/6435
>
> --John
>
> On Tue, Jul 14, 2009 at 7:11 PM, David Flanagan
> <da...@davidflanagan.com>wrote:
>
>
>
> > If I do this:
>
> > $("h1").wrap("<i></i><i></i>")
>
> > Then I get two copies of each of the h1 tags. The return value of the
> > wrap() method does not include the extra copies, however. Is this a bug?
>
> > If so, maybe change this line in wrapAll:
>
> > var wrap = jQuery( html, this[0].ownerDocument).clone();
>
> > To this:
>
> > var wrap = jQuery( html, this[0].ownerDocument).eq(0).clone();
>
> > David
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