How to get a position of a drag clone image using jquery

Asked Mar 28 '11 at 06:03

Active Mar 28 '11 at 06:26

Viewed 560 times

I want position of a drag clone image after drop it on droppable area.

$("#dhtmlgoodies_xpPane .draggable").draggable({ start: function(event, ui) { var Startpos = $(this).position(); $("div#start").text("START: \nLeft: "+ Startpos.left + "\nTop: " + Startpos.top); },

stop: function(event, ui) {
   var Stoppos = $(this).position();
    $("div#stop").text("STOP: \nLeft: "+ Stoppos.left + "\nTop: " + Stoppos.top);
}

}); but it get only image from menu not the clone image.

edited Mar 28 '11 at 06:26

asked Mar 28 '11 at 06:03

Moumita

Did you change the `ID` after cloning cos jquery selector using `ID` s only returns the first match – Jishnu A P Mar 28 '11 at 06:34
no. actually I don't know how to do it. can you refer some code? – Moumita Mar 28 '11 at 06:38
`$("#clonething").clone().attr("id","newID").appendTo("#Whereveryouwant");` – Jishnu A P Mar 28 '11 at 06:41
is it added into the drop section---"$("#working-area").droppable({ here });" – Moumita Mar 28 '11 at 06:46
if have already been using `clone()` in your code add the code to change the `ID` otherwise your problem is not what i think.. – Jishnu A P Mar 28 '11 at 06:55
http://stackoverflow.com/questions/5949172/jquery-draggable-position-clone – Zafer Dec 23 '11 at 09:36

How to get a position of a drag clone image using jquery

0 Answers0