Load jQuery twice and use jQuery.noConflict().
-- Ariel Flesler http://flesler.blogspot.com On 25 mayo, 05:26, Christof Donat <[EMAIL PROTECTED]> wrote: > Hi, > > > > I meant to ask... What is the actual goal here? Can you give a little > > > background? Maybe there is another way to accomplish what you want to do > > > that doesn't rely on cloning the jQuery constructor. > > > I want to have both $ and $D; but in $D, I want to override some > > functions like 'click', 'hover'. Till now, $ and $D gets overwritten > > and so I couldn't use $ or couldn't override $D.click > > Is it absolutelly necessary, that $D has all the other jQuery functionality? > If not just hav the code you need in $D. > > Is it absolutelly necessary, that the functions have the name click() and > hover()? Maybe you can simply add dclick() and dhover to jQuery. > > If both are necessary I'd still stick to Michaels question: Why? Sometimes > there are other solutions to a problem that one doesn't see untill someone > else hits ones nose to it. Tell us why you need it that way and we can think > about alternative solutions. > > Christof
